B. Same Parity Summands
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
Example
inputCopy
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
outputCopy
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
#include<iostream>
using namespace std;
int main()
{
int n,a,b;cin>>n;while(n--){
cin>>a>>b;if(a%2==0){
if(b>a/2){
if(b%2==1)cout<<"NO"<<endl;else if(b<=a){
cout<<"YES"<<endl;for(int i=0;i<b-1;i++)cout<<"1"<<" ";cout<<a-(b-1)<<endl;}else{
cout<<"NO"<<endl;}}else{
cout<<"YES"<<endl;for(int i=0;i<b-1;i++)cout<<"2"<<" ";cout<<a-2*(b-1)<<endl;}}else{
if(b%2==0){
cout<<"NO"<<endl;}else{
if(b>a)cout<<"NO"<<endl;else{
cout<<"YES"<<endl;for(int i=0;i<b-1;i++)cout<<"1"<<" ";cout<<a-(b-1)<<endl;}}}}
}