Codeforces Round #232 (Div. 2) B - On Corruption and Numbers

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B. On Corruption and Numbers

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity — he has not yet decided what profession he wants to get. At school,
he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.

The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission — ni berubleys.
He cannot pay more than ni,
because then the difference between the paid amount and ni can
be regarded as a bribe!

Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri.
The rector also does not carry coins of denomination less than li in
his pocket — because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys,
where li ≤ x ≤ ri (Berland
uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned
out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So,
Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.

Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.

In other words, you are given t requests, each of them contains numbers ni, li, ri.
For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys
using only coins with an integer denomination from li to ri berubleys.
You can use coins of different denominations. Coins of each denomination can be used any number of times.

Input

The first line contains the number of universities t, (1 ≤ t ≤ 1000)
Each of the next t lines contain three space-separated integers: ni, li, ri (1 ≤ ni, li, ri ≤ 109; li ≤ ri).

Output

For each query print on a single line: either "Yes", if Alexey can enter the university, or "No"
otherwise.

Examples

input

2
5 2 3
6 4 5

output

Yes
No

Note

You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.

找到最接近n的f，f为l的倍数，即l*k ≤ n，然后判断，如果r*k≥n的话，即为可行。

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e3+10;
const int maxx=1e6+100;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int main()
{
int t;
LL n,l,r;
scan_d(t);
W(t--)
{
scan_d(n),scan_d(l),scan_d(r);
LL val=n;
n/=l;
if(val<=n*r)
puts("Yes");
else puts("No");
}
}