codeforces 397B

思路： 其中个数最多有n/l，个数最少有n/r，只要n在（n/l*l,n/l*r）之间，最会有答案使得结果为n

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){int t,n,l,r,i,j;scanf("%d",&t); while(t--){scanf("%d%d%d",&n,&l,&r);if(n/l>(n-1)/r) printf("Yes\n");else printf("No\n");}}