Replacing Elements (CodeForces

Replacing Elements

You have an array a1,a2,…,an. All ai are positive integers.

In one step you can choose three distinct indices i, j, and k (i≠j; i≠k; j≠k) and assign the sum of aj and ak to ai, i. e. make ai=aj+ak.

Can you make all ai lower or equal to d using the operation above any number of times (possibly, zero)?

Input
The first line contains a single integer t (1≤t≤2000) — the number of test cases.

The first line of each test case contains two integers n and d (3≤n≤100; 1≤d≤100) — the number of elements in the array a and the value d.

The second line contains n integers a1,a2,…,an (1≤ai≤100) — the array a.

Output
For each test case, print YES, if it’s possible to make all elements ai less or equal than d using the operation above. Otherwise, print NO.

You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example

input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6

output
NO
YES
YES

Note
In the first test case, we can prove that we can’t make all ai≤3.

In the second test case, all ai are already less or equal than d=4.

In the third test case, we can, for example, choose i=5, j=1, k=2 and make a5=a1+a2=2+1=3. Array a will become [2,1,5,3,3].

After that we can make a3=a5+a2=3+1=4. Array will become [2,1,4,3,3] and all elements are less or equal than d=4.

题目还是很好理解的，任选数组中的两个作和替换较大的数，可以先对一串数字进行排序，排序方式从小到大和从大到小都可，sort函数默认排序从小到大。
一旦数组中最大的数即a[n-1]是一个小于或等于d的数，直接输出YES即可，否则运用数组中最小的两个数加和替换最大的数。

（是很简单的题，谨以此篇文章纪念自己会使用sort函数。）

/*by BFU zq2021/1/16 14:13:51*/
#include<bits/stdc++.h>
using namespace std;int main()
{int t,n,d,a[103];cin>>t;while(t--){cin>>n>>d;for (int i=0;i<n;i++){cin>>a[i];}sort(a,a+n);if (a[n-1] > d){int m = a[0] + a[1];a[n-1] = m;if (a[n-1] > d){cout<<"No"<<endl;}else{cout<<"Yes"<<endl;}}else{cout<<"Yes"<<endl;}}return 0;
}

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Replacing Elements (CodeForces - 1473A)

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