HDU 1228 A + B【字符串的处理】

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//对字符串的处理很好

A + B

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12172 Accepted Submission(s): 7118

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>char a[10][10]={"zero","one","two","three","four","five","six","seven","eight","nine"};int fun(char c[]){    //if(strcmp(a[0],c)==0)return 0;//    if(strcmp(a[1],c)==0)return 1;//    if(strcmp(a[2],c)==0)return 2;//    if(strcmp(a[3],c)==0)return 3;//    if(strcmp(a[4],c)==0)return 4;//    if(strcmp(a[5],c)==0)return 5;//    if(strcmp(a[6],c)==0)return 6;//    if(strcmp(a[7],c)==0)return 7;//    if(strcmp(a[8],c)==0)return 8;//    if(strcmp(a[9],c)==0)return 9;  int i;  for(i=0;i<10;i++)  if(strcmp(a[i],c)==0) return i; }    int main(){    char str1[10],str2[10];    int p,q;    while(1)    {        p=q=0;        while(scanf("%s",str1) && strcmp(str1,"+")!=0)        p=p*10 + fun(str1);    while(scanf("%s",str2) && strcmp(str2,"=")!=0)    q=q*10 + fun(str2);if(p==0&&q==0)break;elseprintf("%d\n",p+q);    }        return 0;}

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