标签：字符串

简单计算器

1 + 2
4 + 2 * 5 - 7 / 11
0

3.00
13.36

分析：因为没有括号，所以我们可以按照：如果是数字进模拟队列，判断后面的符号是+-还是*/，如果是乘除将前面的（*=）或（/=）后面的就好，否则讲后面的进队列

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <string>
#include <stdlib.h>
#include <queue>
#include <stack>
using namespace std;
char str[205];
int main(){
    while(gets(str)){
        int len = strlen(str), i, j, k, z;
        double fa, fb;
        double que[250];
        for(i = 0; i < len; i ++)
            if(str[i] != '0') break;
        if(i == len) break;
        //str[len++] = ' '; str[len++] = ' '; str[len] = '\0';
        char temp[205], ca[205];
        int flag = 0;
        for(i = 0, j = 0, k = 0, z = 0; i < len; i ++){
            if(str[i] >= '0'&&str[i] <= '9'){
                temp[j++] = str[i];
            }
            else if(i+1 < len){
                temp[j++] = '\0';
                //printf("%s... ", temp);
                //if(str[i+3] >= '0'&&str[i])
                if(flag){
                    if(flag == 1) que[k-1]*= atol(temp);
                    else que[k-1]/=atol(temp);
                    flag = 0;
                }
                else que[k++] = atol(temp);
                    if(str[i+1] == '*'){
                        flag = 1;
                    }
                    else if(str[i+1] == '/'){
                        flag = 2;
                    }
                    else ca[z++] = str[i+1];
                    i+=2;
                    j = 0;
            }
        }
        temp[j] = '\0';
        if(flag){
            if(flag == 1) que[k-1]*= atol(temp);
            else que[k-1]/=atol(temp);
            flag = 0;
        }
        else que[k++] = atol(temp);
        /*for(i = 0; i < k; i ++) printf("%lf..\n", que[i]);    
        for(i = 0; i < z; i ++) printf("%c..\n", ca[i]);*/
        for(i = 1; i < k; i ++){
            if(ca[i-1] == '+') que[i] += que[i-1];
            else que[i] = que[i-1] - que[i];
        }    
        printf("%.2lf\n", que[k-1]);
    }
    return 0;
}

hdoj 1237 简单计算器 【字符串处理】

标签：字符串

原文地址：http://blog.csdn.net/shengweisong/article/details/40750597