Algorithm
Cirno_9baka has a paper tape with n cells in a row on it. As he thinks that the blank paper tape is too dull, he wants to paint these cells with m kinds of colors. For some aesthetic reasons, he thinks that the i-th color must be used exactly ai times, and for every k consecutive cells, their colors have to be distinct.
Help Cirno_9baka to figure out if there is such a way to paint the cells.
The first line contains a single integer t (1≤t≤10000) — the number of test cases. The description of test cases follows.
The first line of each test case contains three integers n, m, k (1≤k≤n≤109, 1≤m≤105, m≤n). Here n denotes the number of cells, m denotes the number of colors, and k means that for every k consecutive cells, their colors have to be distinct.
The second line of each test case contains m integers a1,a2,⋯,am (1≤ai≤n) — the numbers of times that colors have to be used. It's guaranteed that a1+a2+…+am=n.
It is guaranteed that the sum of m over all test cases does not exceed 105.
For each test case, output "YES" if there is at least one possible coloring scheme; otherwise, output "NO".
You may print each letter in any case (for example, "YES", "Yes", "yes", and "yEs" will all be recognized as positive answers).
NO YES
In the first test case, there is no way to color the cells satisfying all the conditions.
In the second test case, we can color the cells as follows: (1,2,1,2,3,4,3,4,5,6,5,6). For any 2 consecutive cells, their colors are distinct.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl "\n"
#define debug(n) cout<<(n)<<endl;
#define pb push_back
const ll INF = 2e18 + 99;
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin>>t;
while(t--){
int n, m, k;
cin>>n>>m>>k;
int seg;
if (n % k != 0){
seg = n/k + 1;
}
else{
seg = n /k;
}
int x, count = 0;
bool check = true;
for(int i = 0; i < m; i++){
cin>>x;
if(x > seg){
check = false;
}
if(x == seg){
count++;
}
}
if(n % k != 0 && count > n % k){
check = false;
}
if(check){
cout<<"YES"<<endl;
}
else{
cout<<"NO"<<endl;
}
}
}Input
12 6 2
1 1 1 1 1 7
12 6 2
2 2 2 2 2 2
Output
YES
Demonstration
Codeforcess Solution 1774-B B. Coloring ,C++, Java, Js and Python,1774-B,Codeforcess Solution