hdu 1228 A+B【字符串处理】

要用gets输入    纠结了一上午！！ 要细心啊！注意新的字符串后面的'\0';

 

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

39096

#include<stdio.h>#include<string.h>const char *s[10] = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" };int main(){char str[100], a[4][50], b[50], temp[10];int aa, bb, i, j, k;while( gets(str) ){i = k = j = 0;int flag = 0;aa = bb= 0;while( str[i] != '=' ){if( str[i]>='a'&&str[i]<='z' )a[j][k++] = str[i];else if( str[i+1] != ' '&&str[i+1] != '+' ){a[j][k] = '\0'; //要注意这处++j; k = 0;} if( str[i] == '+' )flag = j;++i;}a[j][k] = '\0';//这处for( i = 0; i <= flag; i ++ ){for( k = 0; k < 10; k ++ ){if(  strcmp( a[i], s[k] ) == 0 ){aa = aa*10+k;break;}}}for( i = flag+1; i < j; i ++ )for( k = 0; k < 10; k ++ )if( strcmp( a[i], s[k] ) == 0 ){bb = bb*10+k;break;}if( aa == 0&&bb==0 )break;printf( "%d\n", aa+bb );}return 0;}