Codeforces Round #640 (Div. 4)B. Same Parity Summands

Problem description

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

input
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

题意：找到全为奇数或者全为偶数的k个数，使他们的和为n

#include 
#include 
using namespace std;
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, k;
        cin >> n >> k;
        int n1 = n - (k - 1);
        if (n1 > 0 && n1 % 2 == 1)
        {
            cout << "YES" << endl;
            for (int i = 0; i < k - 1; ++i)
                cout << "1 ";
            cout << n1 << endl;
            continue;
        }
        int n2 = n - 2 * (k - 1);
        if (n2 > 0 && n2 % 2 == 0)
        {
            cout << "YES" << endl;
            for (int i = 0; i < k - 1; ++i)
                cout << "2 ";
            cout << n2 << endl;
            continue;
        }
        cout << "NO" << endl;
    }
    return 0;
}