D. Game With Array

// Problem: D. Game With Array
// Contest: Codeforces - Codeforces Round #643 (Div. 2)
// URL: https://codeforces.com/problemset/problem/1355/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-03-01 16:35:40
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
void solve() {
    int n, s;
    cin >> n >> s;
    if (n == 1) {
        if (s == 1) cout << "NO\n";
        else {
            cout << "YES\n";
            cout << s << endl;
            cout << 1 << endl;
        }
    }
    else {
        int k = s / n;
        if (k == 1) {
            puts("NO");
        }
        else {
            puts("YES");
            int r = s % n;
            for (int i = 0; i < n - 1; i++)
                cout << k << " ";
            cout << k + r << endl;
            cout << k - 1 << endl;
        }
    }
}
int main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

先看小的，比如1，然后尽可能均匀发分配，只要存在1那么就无法赢。否则能赢