A + B

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

Total Submission(s): 12192    Accepted Submission(s): 7132

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

Source

浙大计算机研究生复试上机考试-2005年

#include <stdio.h>
#include <string.h>

#define maxn 100

const char *sam[] = {"zero", "one", "two", "three", "four",
					"five", "six", "seven", "eight", "nine"};
char buf[maxn];

int calVal(char str[]) {
	for(int i = 0; i < 10; ++i)
		if(!strcmp(buf, sam[i]))
			return i;
}

int main() {
	int mode = 0, a[2];
	a[0] = a[1] = 0;
	while(scanf("%s", buf) == 1) {
		if(!strcmp(buf, "+")) {
			++mode; continue;
		} else if(!strcmp(buf, "=")) {
			if(a[0] + a[1] == 0) break;
			printf("%d\n", a[0] + a[1]);
			mode = a[0] = a[1] = 0;
			continue;
		}
		a[mode] = a[mode] * 10 + calVal(buf);
	}
	return 0;
}