HDU-1228-A+B（字符串处理）

hdu 1228 A+B

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

思路：关键就是字符串的处理，怎么把每个数给分离出来，用scanf一个一个的读入，用change()函数给转换成具体的数字

[code]#include<cstdio>
#include<cstring>

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

int change(char s[]){
    for(int i = 0; i < 10; i++)
        if(!strcmp(a[i], s))
            return i;
}

int main()
{
    char str[10], num[10];
    while(1){
        int a = 0, b = 0;
        while(scanf("%s", str) && strcmp(str, "+")){
            a = a*10+change(str);
        }
        while(scanf("%s", str) && strcmp(str, "=")){
            b = b*10+change(str);
        }
        if(!(a+b))  break;
        printf("%d\n", a+b);
    }
    return 0;
}