HDU 1228 A + B

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A + B

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 2 Accepted Submission(s) : 1

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

Source

浙大计算机研究生复试上机考试-2005年

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char map[12][10]={"zero","one","two","three","four","five","six","seven","eight","nine"};int init(char ch[]){int i;for(i=0;i<=10;i++){if(!strcmp(map[i],ch))break;}return i;}int main(){char s[11];int a,b;while(scanf("%s",s)!=EOF){a=b=0;while(strcmp(s,"+")!=0){a=a*10+init(s);scanf("%s",s);}scanf("%s",s);while(strcmp(s,"=")!=0){b=b*10+init(s);scanf("%s",s);}if(a+b==0)break;else printf("%d\n",a+b);}return 0;}

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