Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11043 Accepted Submission(s): 6376
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
#include<stdio.h> #include<string.h> char str[100]; char s[100][6]; char flag[11][6]={"zero","one","two","three","four","five","six","seven","eight","nine","ten"}; int main() { int i,j,k,u,n,m,v1,v2,num; while(gets(str)) { n=strlen(str); k=0;u=0; for(i=0;i<n;i++) { if(str[i]!=' ') s[k][u++]=str[i]; else { k++;u=0; } } if(strcmp(s[0],"zero")==0&&strcmp(s[2],"zero")==0) {break;} else { v1=0;v2=0;num=1; for(i=0;i<k;i++) { if(strcmp(s[i],"+")!=0) { for(j=0;j<10;j++) { if(strcmp(s[i],flag[j])==0) { if(num==1) v1=v1*10+j; else v2=v2*10+j; } } } else { num=2; } } //printf("%d+%d=",v1,v2); printf("%d\n",v1+v2); memset(str,0,sizeof(str)); memset(s,0,sizeof(s)); } } return 0; }