Codeforces Round #232 (Div. 2), problem: (B) On Corruption and Numbers solution

Codeforces Round #232 (Div. 2), problem: (B) On Corruption and Numbers: http://codeforces.ru/contest/397/problem/B

Codeforces Round #232 (Div. 2), problem: (B) On Corruption and Numbers editorial:

Div2-B

First of all, calculate how many L's we can bring so that the result will not exceed N. It's .

Now, if R·K ≥ N we may increase any of K numbers by 1. At some moment sum will be equal to N becase sum of K R's is greater thanN. So answer in this case is YES.

If R·K < N, we can't get K or less numbers because their sum is less than N. But we can't get more than K numbers too, because their sum is more than N. So answer is NO.

Complexity: O(1) for every query.
Overall complexity: O(t).

Codeforces Round #232 (Div. 2), problem: (B) On Corruption and Numbers solution: http://ideone.com/r2fZHe

#include <iostream>
#include <cstdio>
using namespace std;

int main() {
    int t, n, l, r;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d%d", &n, &l, &r);
        if(n%l==0 || n%r==0 || (n%l)%r==0 || (n%r)%l==0) {
            printf("Yes\n");
            continue;
        }
        if(r*(n/r+1)>n && l*(n/r+1)<n) {
            printf("Yes\n");
            continue;
        }
        printf("No\n");
    }
    return 0;
}