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;

}