HDOJ 1228 A + B

 

A + B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10194    Accepted Submission(s): 5832

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

浙大计算机研究生复试上机考试-2005年

 

 

Recommend

JGShining

 

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

char a[10][10]={"zero","one","two","three","four","five","six",
"seven","eight","nine"};

int main()
{
    int end1=0,end2=0;
while(true)
{
    end1=0;end2=0;
    int A=0,B=0;
    char nu[10];
    int i;
while(scanf("%s",nu))
{
    if(strcmp(nu,"+")==0)  break;
    for(i=0;i<10;i++)
    {
        if(strcmp(nu,a[i])==0)
            break;
    }
    A=A*10+i;
}

while(scanf("%s",nu))
{
    if(strcmp(nu,"=")==0)  break;
    for(i=0;i<10;i++)
    {
        if(strcmp(nu,a[i])==0)
            break;
    }
    B=B*10+i;
}
//cout<<A<<"..."<<B<<endl;
if(A==0) end1=1;
if(B==0) end2=1;
if(end1==1&&end2==1)
    break;
else
    cout<<A+B<<endl;
}

    return 0;
}