多项式加减乘
描述： 多项式加减乘
有A，B，C…… n 个关于x的个多项式，求一个关于A，B，C……的多项式的结果
多项式的每一项之间用空格隔开，每一项的系数，x，指数之间也用空格隔开，系数为0的项不输出
输入： 基于多项式A，B，C……的运算表达式
关于x的多项式A
关于x的多项式B
关于x的多项式C
…
输出： 表达式的运算结果
含有A，B，C……表达式的运算结果，保留两位小数。
输出多项式表达式按指数降序排序
输入样例： (A+B)*(A-B)
A 1 x 2.1 3 x -3.09
B 2 x 1 2 x 0
输出样例： 1.00 x 4.20 -4.00 x 2.00 -8.00 x 1.00 -4.00 x 0.00 6.00 x -0.99 9.00 x -6.18

[解决办法]

C/C++ code

//表达式求值#include <stdio.h>#include <malloc.h>#include <string.h>/**功能:根据运算符计算*参数:a, b参与运算的数, ch运算符*返回值:计算结果,操作符错误则返回0*/int cal(int a, char ch, int b){    switch(ch)    {    case '+':        return a+b;        break;    case '-':        return a-b;        break;    case '*':        return a*b;        break;    case '/':        return a/b;        break;    }    return 0;}/**功能:计算表达式的值(用数组模拟栈)*参数:表达式字符串*返回值:计算结果*/int evaluateExpression(char *str){    int i = 0, result, numSub = 0, operSub = 0;    int tmp, len = strlen(str);    int *operand = (int*)malloc(sizeof(int)*len);    char *operat = (char*)malloc(sizeof(char)*len);    while(str[i] != '\0')    {        switch(str[i])        {        case '+':            while(operSub > 0 && operat[operSub-1] != '(')            {                printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]);                operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]);                printf("%d\n", operand[numSub-2]);                --numSub;                --operSub;            }            operat[operSub++] = '+';            break;        case '-':            while(operSub > 0 && operat[operSub-1] != '(')            {                printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]);                operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]);                printf("%d\n", operand[numSub-2]);                --numSub;                --operSub;            }            operat[operSub++] = '-';            break;        case '*':            if(str[i+1] >= '0' && str[i+1] <= '9')            {                tmp = 0;                while(str[i+1] >= '0' && str[i+1] <= '9')                {                    tmp = tmp * 10 + str[i+1] - '0';                    ++i;                }                --i;                printf("%d * %d = ", operand[numSub-1], tmp);                operand[numSub-1] = cal(operand[numSub-1], '*', tmp);                printf("%d\n", operand[numSub-1]);                ++i;            }            else                operat[operSub++] = '*';            break;        case '/':            if(str[i+1] >= '0' && str[i+1] <= '9')            {                tmp = 0;                while(str[i+1] >= '0' && str[i+1] <= '9')                {                    tmp = tmp * 10 + str[i+1] - '0';                    ++i;                }                --i;                printf("%d / %d = ", operand[numSub-1], tmp);                operand[numSub-1] = cal(operand[numSub-1], '/', tmp);                printf("%d\n", operand[numSub-1]);                ++i;            }            else                operat[operSub++] = '/';            break;        case '(':            operat[operSub++] = '(';            break;        case ')':            while(operat[operSub-1] != '(')            {                printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]);                operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]);                printf("%d\n", operand[numSub-2]);                --numSub;                --operSub;            }            --operSub;            break;        default:            tmp = 0;            while(str[i] >= '0' && str[i] <= '9')            {                tmp = tmp * 10 + str[i] - '0';                ++i;            }            --i;            operand[numSub++] = tmp;            break;        }        ++i;    }    while(numSub > 1 && operSub >= 1)    {        printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]);        operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]);        printf("%d\n", operand[numSub-2]);        --numSub;        --operSub;    }    result = operand[numSub-1];    free(operand);    free(operat);    return result;}int main(){    char *str = "225/15-20+(4-3)*2";    int result;    printf("计算过程:\n");    result = evaluateExpression(str);    printf("计算结果:result = %d\n", result);    return 0;}