转化逆波兰式为常规表达式

题目来自《程序设计导引及在线实践》9.4思考题 转化逆波兰式为正常的表达式

#include <stdio.h>#include <string.h>//输入样例：//* + 11.0 12.0 + 24.0 35.0//输出样例//((11.0+12.0)*(24.0+35.0))char* exp2(){char a[100];char buff1[256];char buff2[256];char buff_sum[256];scanf("%s",a);switch (a[0]){case '+':strcpy(buff1,exp2());strcpy(buff2,exp2());sprintf(buff_sum,"(%s+%s)",buff1,buff2);return buff_sum;case '-': strcpy(buff1,exp2());strcpy(buff2,exp2());sprintf(buff_sum,"(%s-%s)",buff1,buff2);return buff_sum;case '*':strcpy(buff1,exp2());strcpy(buff2,exp2());sprintf(buff_sum,"(%s*%s)",buff1,buff2);return buff_sum;case '/':strcpy(buff1,exp2());strcpy(buff2,exp2());sprintf(buff_sum,"(%s/%s)",buff1,buff2);return buff_sum;default:return (a);}}int main(int argc, char* argv[]){char res[256];strcpy(res,exp2());printf("%s",res);return 0;}