NYOJ64 鸡兔同笼

原题链接


#include <stdio.h>

int main(){
	int t, n, m,a ,b;
	scanf("%d", &t);
	while(t--){
		scanf("%d%d", &n, &m);
		a = (m - 2 * n) / 2; //rabbit
		b = n - a;
		if(a < 0 || b < 0 || m & 1)
			printf("No answer\n");
		else printf("%d %d\n", b, a);
	}
	return 0;
}


posted on 2014-03-17 17:22  长木Qiu  阅读(147)  评论(0编辑  收藏  举报