题目描述

You are given two positive integers n (1≤n≤109) and k (1≤k≤100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).

In other words, find a1,a2,…,ak such that all ai>0, n=a1+a2+…+ak and either all ai are even or all ai are odd at the same time.

If such a representation does not exist, then report it.

Input
The first line contains an integer t (1≤t≤1000) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n (1≤n≤109) and k (1≤k≤100).

Output
For each test case print:

YES and the required values ai, if the answer exists (if there are several answers, print any of them);
NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.

样例

8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9

YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120

这个题要求我们把n分解成k个偶数或者k个奇数之和
首先我们可以把 k个数看成 k-1个数和第k个数
如果我们要分成k个奇数 我们可以分解成 k-1个1 和一个n-(k-1)只要判断 n-(k-1)是不是奇数即可
同理
如果我们要分成k个偶数 我们可以分解成 k-1个2 和一个n-2(k-1) 只要判断 n-2(k-1)是不是偶数即可

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        int n,k;
        cin>>n>>k;
        int n1=n-(k-1);
        int n2=n-2*(k-1);
        if(n1>0&&n1%2==1)
        {
            cout<<"YES"<<endl;
            for(int i=1;i<=k-1;i++)
            cout<<1<<' ';
            cout<<n1<<endl;
        }
        else if(n2>0&&n2%2==0)
        {
            cout<<"YES"<<endl;
            for(int i=1;i<=k-1;i++)
            cout<<2<<' ';
            cout<<n2<<endl;
        }
        else{
            cout<<"NO"<<endl;
        }
    }
}