杭电oj~1228 map=_=

A + B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15387    Accepted Submission(s): 9162


Problem Description
读入两个小于100的正整数A和B,计算A+B.
需要注意的是:A和B的每一位数字由对应的英文单词给出.
 
Input
测试输入包含若干测试用例,每个测试用例占一行,格式为"A + B =",相邻两字符串有一个空格间隔.当A和B同时为0时输入结束,相应的结果不要输出. 
 
Output
对每个测试用例输出1行,即A+B的值.
 
Sample Input
one + two = three four + five six = zero seven + eight nine = zero + zero =
 
Sample Output
3 90 96
 
Source
 
 
 
 
#include<iostream>
#include<map>
#include<cstring>
using namespace std;

map<string,int>m;

int main()
{
    m["zero"]=0;
    m["one"]=1;
    m["two"]=2;
    m["three"]=3;
    m["four"]=4;
    m["five"]=5;
    m["six"]=6;
    m["seven"]=7;
    m["eight"]=8;
    m["nine"]=9;
    m["+"]=10;
    m["="]=11;
    int cnt[2]={0};
    char a[20];
    int k=0;int z=0;
    int sum[2]={0};
    while(cin>>a)
    {
        int d=m[a];
        if(d<10) cnt[k++]=d;
        if(d==10)
        {
            if(k==1) sum[0]=cnt[0];
            else if(k==2) sum[0]=10*cnt[0]+cnt[1];
            k=0;
        }
        if(d==11)
        {
            if(k==1) sum[1]=cnt[0];
            else if(k==2) sum[1]=10*cnt[0]+cnt[1];
            k=0;
            if(sum[0]+sum[1]==0) break;
            else cout<<sum[0]+sum[1]<<endl;
        }
    }
    return 0;

}

  

 
原文地址:https://www.cnblogs.com/hellohacker/p/5749371.html