134 lines
2.5 KiB
C++
134 lines
2.5 KiB
C++
#include <bits/stdc++.h>
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// https://codeforces.com/blog/entry/96344
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#pragma GCC optimize("O2,unroll-loops")
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#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
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using namespace std;
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template <typename... Args>
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void print(std::string const &str, Args &&...args) {
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std::cout << std::vformat(str,
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// make_format_args binds arguments to const
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std::make_format_args(args...));
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}
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template <typename T>
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void print(T const &t) {
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std::cout << t;
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}
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template <std::ranges::range T>
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void print(T const &t) {
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if constexpr (std::is_convertible_v<T, char const *>) {
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std::cout << t << '\n';
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} else {
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for (auto const &e : t) {
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std::cout << e << ' ';
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}
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std::cout << '\n';
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}
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}
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template <typename... Args>
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void println(std::string const &str, Args &&...args) {
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print(str, std::forward<Args>(args)...);
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cout << '\n';
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}
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template <typename T>
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void println(T const &t) {
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print("{}\n", t);
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}
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template <std::ranges::range T>
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void println(T const &t) {
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cout << t << '\n';
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}
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void println() {
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std::cout << '\n';
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}
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template <typename T>
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T MAX() {
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return std::numeric_limits<T>::max();
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}
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template <typename T>
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T MIN() {
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return std::numeric_limits<T>::min();
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}
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#define ff first
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#define ss second
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#define eb emplace_back
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#define ll long long
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#define ld long double
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#define vec vector
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#define all(x) (x).begin(), (x).end()
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#define rall(x) (r).rbegin(), (x).rend()
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#define sz(x) static_cast<int>((x).size())
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#define FOR(a, b, c) for (int a = b; a < c; ++a)
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mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
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#define randint(a, b) uniform_int_distribution(a, b)(rng)
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void YES() {
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cout << "YES\n";
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}
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void NO() {
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cout << "NO\n";
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}
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#ifdef LOCAL
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#define dbg(x) cout << __LINE__ << ": " << #x << "=<" << (x) << ">\n";
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#else
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#define dbg(x)
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#endif
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static constexpr int MOD = 1e9 + 7;
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void solve() {
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/*
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n, k; SUM ai = n, same parity
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n odd: k even -> NO; k odd -> YES
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n even: k even: odd #s -> NO, only even
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*/
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int n, k;
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cin >> n >> k;
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if (k > n) {
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NO();
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return;
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}
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if ((n - 2 * (k - 1)) % 2 == 0 && (n - 2 * (k - 1)) > 0) {
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YES();
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FOR(i, 1, k)
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cout << 2 << ' ';
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cout << n - 2 * (k - 1) << '\n';
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} else if ((n - k + 1) & 1) {
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YES();
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FOR(i, 1, k)
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cout << 1 << ' ';
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cout << n - k + 1 << '\n';
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} else
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NO();
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}
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int main() {
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cin.tie(nullptr)->sync_with_stdio(false);
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int t = 1;
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cin >> t;
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while (t--) {
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solve();
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}
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return 0;
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}
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