NYOJ664 数字整除

原题链接

（a-b）%c = (a%c - b%c)%c

#include <stdio.h>
#include <string.h>
char str[105];

int main(){
	int c, len, a, i;
	while(scanf("%s", str), str[0] != ‘0‘){
		len = strlen(str);
		if(len == 1){
			printf("0\n");
			continue;
		}
		c = str[len - 1] - ‘0‘;
		str[len - 1] = ‘\0‘;
		if(len < 10){					
			sscanf(str, "%d", &a);
			a -= c * 5;
			printf("%d\n", a % 17 ? 0 : 1);
			continue;
		}
		//大数取余
		for(i = a = 0; i != len - 1; ++i){
			a = (a * 10 + str[i] - ‘0‘) % 17;
		}
		c = c * 5 % 17;
		printf("%d\n", (a - c) % 17 ? 0 : 1);
	}
	return 0;
}

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NYOJ664 数字整除

原文：http://blog.csdn.net/chang_mu/article/details/20801375