题目描述
明明进了中学之后,学到了代数表达式。有一天,他碰到一个很麻烦的选择题。这个题目的题干中首先给出了一个代数表达式,然后列出了若干选项,每个选项也是一个代数表达式,题目的要求是判断选项中哪些代数表达式是和题干中的表达式等价的。
这个题目手算很麻烦,因为明明对计算机编程很感兴趣,所以他想是不是可以用计算机来解决这个问题。假设你是明明,能完成这个任务吗?
这个选择题中的每个表达式都满足下面的性质:
1. 表达式只可能包含一个变量‘a’。
2. 表达式中出现的数都是正整数,而且都小于10000。
3. 表达式中可以包括四种运算‘ ’(加),‘-’(减),‘*’(乘),‘^’(乘幂),以及小括号‘(’,‘)’。小括号的优先级最高,其次是‘^’,然后是‘*’,最后是‘ ’和‘-’。‘ ’和‘-’的优先级是相同的。相同优先级的运算从左到右进行。(注意:运算符‘ ’,‘-’,‘*’,‘^’以及小括号‘(’,‘)’都是英文字符)
4. 幂指数只可能是1到10之间的正整数(包括1和10)。
5. 表达式内部,头部或者尾部都可能有一些多余的空格。
下面是一些合理的表达式的例子:
((a^1) ^ 2)^3,a*a a-a,((a a)),9999 (a-a)*a,1 (a -1)^3,1^10^9……
分析:表达式运算,一般都是用一个操作数栈,一个操作符栈,依据操作符的优先级逐级运算
本题的麻烦在于有一个变量,字符串操作起来就比较繁琐。
于是乎,就将多项式的每一项抽象为一个数据结构,因为每个项都可以由一个系数和指数唯一确定;那么一个多项式就可以表示为一组【系数 指数】的集合;
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#include <stdio.h>
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#include <stdlib.h>
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#include <string.h>
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#define max_exponent 100
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#define max_polynomial 100
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typedef struct polynomial_node {
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unsigned int number;
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short *coeff;
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unsigned short *exp;
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} polynomial_t;
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polynomial_t* generate_from_list(short *exp_coeff, int len)
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{
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polynomial_t *ret = null;
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short *coeffs;
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unsigned short *exps;
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int cnt = 0;
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int i, j;
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for (i = 0; i < len; i)
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{
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if (exp_coeff[i] != 0)
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{
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cnt;
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}
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}
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if (cnt == 0)
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{
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cnt = 1;// return 0
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}
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ret = (polynomial_t *)calloc(sizeof(polynomial_t), 1);
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if (null == ret)
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{
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return null;
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}
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coeffs = (short *)calloc(sizeof(short), cnt);
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exps = (unsigned short *)calloc(sizeof(unsigned short), cnt);
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if (null == coeffs || null == exps)
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{
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return null;
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}
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j = 0;
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for (i = 0; i < len; i)
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{
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if (exp_coeff[i] != 0)
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{
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coeffs[j] = exp_coeff[i];
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exps[j] = i;
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}
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}
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ret->number = cnt;
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ret->coeff = coeffs;
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ret->exp = exps;
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return ret;
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}
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polynomial_t* polynomial_multip(polynomial_t *p1, polynomial_t *p2)
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{
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short exp_coeff[max_exponent] = { 0 };
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int i, j;
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for (i = 0; i < p2->number; i)
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{
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for (j = 0; j < p1->number; j)
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{
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exp_coeff[p1->exp[j] p2->exp[i]] = p1->coeff[j] * p2->coeff[i];
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}
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}
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return generate_from_list(exp_coeff, max_exponent);
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}
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polynomial_t* polynomial_add(polynomial_t *p1, polynomial_t *p2)
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{
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short exp_coeff[max_exponent] = { 0 };
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int i, j;
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for (i = 0; i < p1->number; i)
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{
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exp_coeff[p1->exp[i]] = p1->coeff[i];
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}
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for (i = 0; i < p2->number; i)
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{
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exp_coeff[p2->exp[i]] = p2->coeff[i];
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}
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return generate_from_list(exp_coeff, max_exponent);
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}
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polynomial_t* polynomial_sub(polynomial_t *p1, polynomial_t *p2)
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{
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short exp_coeff[max_exponent] = { 0 };
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int i, j;
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for (i = 0; i < p1->number; i)
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{
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exp_coeff[p1->exp[i]] = p1->coeff[i];
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}
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for (i = 0; i < p2->number; i)
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{
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exp_coeff[p2->exp[i]] -= p2->coeff[i];
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}
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return generate_from_list(exp_coeff, max_exponent);
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}
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void polynomial_free(polynomial_t *p)
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{
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if (!p)
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{
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return;
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}
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if (p->exp)
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{
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free(p->exp);
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}
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if (p->coeff)
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{
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free(p->coeff);
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}
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free(p);
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}
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void polynomial_print(polynomial_t *p)
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{
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int i;
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char sym;
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char py[16];
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if (!p)
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{
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return;
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}
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printf("polynomial<%p>, total %d:\n", p, p->number);
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for (i = 0; i < p->number; i)
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{
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memset(py, 0, 16);
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if (p->coeff[i] < 0)
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{
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if (p->exp[i] == 0)
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{
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sprintf(py, "%d", p->coeff[i]);
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}
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else if (p->exp[i] == 1)
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{
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if (p->coeff[i] == -1)
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{
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sprintf(py, "-x");
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}
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else
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sprintf(py, "%dx", p->coeff[i]);
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}
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else
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{ if (p->coeff[i] == -1)
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{
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sprintf(py, "-x^%d", p->exp[i]);
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}
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else
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sprintf(py, "%dx^%u", p->coeff[i], p->exp[i]);
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}
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}
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else
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{
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char* tmp = py;
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if (i > 0)
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{
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tmp[0] = '';
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tmp;
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}
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if (p->exp[i] == 0)
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{
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sprintf(tmp, "%d", p->coeff[i]);
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}
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else if (p->exp[i] == 1)
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{
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if (p->coeff[i] == 1)
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{
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sprintf(tmp, "x");
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}
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else
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sprintf(tmp, "%dx", p->coeff[i]);
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}
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else
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{
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if (p->coeff[i] == 1)
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{
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sprintf(tmp, "x^%d", p->exp[i]);
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}
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else
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sprintf(tmp, "%dx^%d", p->coeff[i], p->exp[i]);
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}
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}
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printf("%s", py);
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}
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printf("\n");
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}
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polynomial_t* polynomial_power(polynomial_t *p1, polynomial_t *p2)
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{
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short exp_coeff[max_exponent] = { 0 };
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int i, j;
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int exp = 0;
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polynomial_t *ret;
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polynomial_t *tmp;
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if (p2->number != 1)
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{
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printf("polynomial power only support number\n");
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return null;
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}
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exp = p2->coeff[0];
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if (exp == 0)
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{
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exp_coeff[0] = 1;
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return generate_from_list(exp_coeff, max_exponent);
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}
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else if (exp < 0)
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{
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printf("polynomial power only support positive exponal\n");
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return null;
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}
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if (exp == 1)
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{
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return p1;
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}
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ret = p1;
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tmp = null;
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while (exp > 1)
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{
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ret = polynomial_multip(ret, p1);
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if (tmp)
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{
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polynomial_free(tmp);
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}
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tmp = ret;
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exp--;
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}
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return ret;
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}
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char* parse_polynomial(char *str, polynomial_t **ptr)
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{
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char *cur = str;
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polynomial_t *ret = null;
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short exp_coeff[2] = { 0 };
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short coeff = 0;
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if (*cur == 'x')
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{
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exp_coeff[1] = 1;
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ret = generate_from_list(exp_coeff, 2);
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*ptr = ret;
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return cur;
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}
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while (*cur >= '0' && *cur <= '9')
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{
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coeff = coeff * 10 (*cur - '0');
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cur;
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}
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exp_coeff[0] = coeff;
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ret = generate_from_list(exp_coeff, 2);
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*ptr = ret;
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return cur;
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}
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polynomial_t* polynomial_comput(char *str)
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{
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char symstack[max_polynomial] = { 0, };
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int symhead = -1;
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polynomial_t *pstack[max_polynomial] = { null };
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int phead = -1;
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polynomial_t *opt1;
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polynomial_t *opt2;
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polynomial_t *ret;
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char optch;
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char *cur;
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if (!str)
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{
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return null;
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}
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cur = str;
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while (*cur != '\0')
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{
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if (*cur == 'x' || (*cur >= '0' && *cur <= '9'))
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{
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cur = parse_polynomial(cur, &pstack[phead]);
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continue;
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}
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else
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{
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switch (*cur)
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{
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case '(': // 左括号直接入栈
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symstack[symhead] = *cur;
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break;
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case '^': // 查看符号栈顶是否为'^',如果是 弹出运算,其余的直接压栈
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if (symhead >= 0 && symstack[symhead] == '^')
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{
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optch = symstack[symhead--];
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opt2 = pstack[phead--];
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opt1 = pstack[phead--];
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ret = polynomial_power(opt1, opt2);
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pstack[phead] = ret;
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symstack[symhead] = *cur;
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polynomial_free(opt2);
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polynomial_free(opt1);
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}
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else
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{
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symstack[symhead] = *cur;
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}
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break;
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case '*': // 符号栈顶为'^' 或'*'则运算,否则压栈
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if (symhead >= 0 && (symstack[symhead] == '^' || symstack[symhead] == '*'))
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{
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optch = symstack[symhead--];
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opt2 = pstack[phead--];
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opt1 = pstack[phead--];
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if (optch == '^')
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ret = polynomial_power(opt1, opt2);
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else
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ret = polynomial_multip(opt1, opt2);
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pstack[phead] = ret;
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symstack[symhead] = *cur;
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polynomial_free(opt2);
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polynomial_free(opt1);
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}
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else
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{
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symstack[symhead] = *cur;
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}
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break;
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case '':
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case '-':
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if (symhead >= 0 & symstack[symhead] != '(')
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{
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optch = symstack[symhead--];
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opt2 = pstack[phead--];
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opt1 = pstack[phead--];
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if (optch == '^')
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ret = polynomial_power(opt1, opt2);
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else if (optch == '*')
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{
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ret = polynomial_multip(opt1, opt2);
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}
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else if (optch == '')
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{
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ret = polynomial_add(opt1, opt2);
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}
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else if (optch == '-')
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{
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ret = polynomial_sub(opt1, opt2);
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}
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pstack[phead] = ret;
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symstack[symhead] = *cur;
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polynomial_free(opt2);
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polynomial_free(opt1);
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}
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else
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symstack[symhead] = *cur;
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break;
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case ')': // 如果是右括号,则一直运算直到找到左括号
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while (symhead >= 0 && symstack[symhead] != '(')
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{
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optch = symstack[symhead--];
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opt2 = pstack[phead--];
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opt1 = pstack[phead--];
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if (optch == '^')
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ret = polynomial_power(opt1, opt2);
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else if (optch == '*')
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{
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ret = polynomial_multip(opt1, opt2);
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}
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else if (optch == '')
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{
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ret = polynomial_add(opt1, opt2);
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}
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else if (optch == '-')
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{
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ret = polynomial_sub(opt1, opt2);
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}
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pstack[phead] = ret;
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}
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if (symstack[symhead] != '(')
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{
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printf("miss match of ')'\n");
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return null;
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}
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symhead--;
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break;
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default:
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break;
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}
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cur;
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}
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}
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while (symhead >= 0)
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{
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optch = symstack[symhead--];
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opt2 = pstack[phead--];
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opt1 = pstack[phead--];
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if (optch == '^')
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{
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ret = polynomial_power(opt1, opt2);
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}
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else if (optch == '*')
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{
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ret = polynomial_multip(opt1, opt2);
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}
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else if (optch == '')
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{
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ret = polynomial_add(opt1, opt2);
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}
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else if (optch == '-')
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{
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ret = polynomial_sub(opt1, opt2);
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}
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pstack[phead] = ret;
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polynomial_free(opt2);
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polynomial_free(opt1);
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}
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if (phead != 0)
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{
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printf("error : %d\n", phead);
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polynomial_print(pstack[phead]);
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return null;
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}
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return pstack[0];
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}
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int main(int argc, char **argv)
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{
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polynomial_t *pyn;
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char src[32] = "(x 1)^2^2-(x-x)^3 ((x-1)*(x-1))";
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pyn = polynomial_comput(src);
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polynomial_print(pyn);
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return 0;
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}
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