SPOJ ONP 简单中缀变后缀 (13.07.07)

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Problem code: ONP

Transform the algebraic expression with brackets into RPN form (Reverse Polish Notation). Two-argument operators: +, -, *, /, ^ (priority from the lowest to the highest), brackets ( ). Operands: only letters: a,b,...,z. Assume that there is only one RPN form (no expressions like a*b*c).

Input

t [the number of expressions <= 100]expression [length <= 400][other expressions]

Text grouped in [ ] does not appear in the input file.

Output

The expressions in RPN form, one per line.

Example

Input:3(a+(b*c))((a+b)*(z+x))((a+t)*((b+(a+c))^(c+d)))Output:abc*+ab+zx+*at+bac++cd+^*此题不论操作符优先级, 一律加括号, 所以好做, 步骤如下:一. 遇到操作数, 直接输出;二. 遇到操作符(除了')'以外), 压入栈中;三. 遇到')', 将与之配对的'('之前的的操作符输出;四. 读入结束, 将栈中操作符全部输出~代码如下:#include<stdio.h>#include<stdlib.h>#include<string.h>char stack1[400];char stack2[400];int top1, top2;void deal(char *p) {    int i, len;    memset(stack1, 0, sizeof(stack1));    memset(stack2, 0, sizeof(stack2));    top1 = 0;    top2 = 0;    len = strlen(p);    for(i = 0; i < len; i++) {        if(p[i] >= 'a' && p[i] <= 'z')            printf("%c", p[i]);        if(p[i] == '(')            stack1[top1++] = p[i];        if(p[i] == '+' || p[i] == '-' || p[i] == '*' || p[i] == '/' || p[i] == '^')            stack2[top2++] = p[i];        if(p[i] == ')' && stack1[top1-1] == '(') {            printf("%c", stack2[--top2]);            top1--;        }    }}int main() {    int t;    scanf("%d", &t);    getchar();    while(t--) {        char str[600];        gets(str);        deal(str);        printf("\n");    }}