Codeforces Round #643 (Div. 2) D. Game With Array

D. Game With Array

题目链接-D. Game With Array


题目大意
构造一个长度为$n$n的序列，并且$n$n个数的和为$S$S，问能不能找到一个数$k∈[1,S]$k∈[1,S]，使得数组里找不出一个子序列的和为$k$k或者$n-k$n−k

解题思路
$贪心$贪心

• 取$k=1$k=1，数组中的前$n-1$n−1个元素都设为$2$2，第$n$n个元素为$S-(n-1)×2$S−(n−1)×2 ，只要第$n$n个元素不等于$1$1 即可，前提是n<=s/2，否则就输出NO
• 具体操作见代码

附上代码

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
#define int long long
#define lowbit(x) (x &(-x))
#define endl '\n'
using namespace std;
const int INF=0x3f3f3f3f;
const int dir[4][2]={-1,0,1,0,0,-1,0,1};
const double PI=acos(-1.0);
const double e=exp(1.0);
const double eps=1e-10;
const int M=1e9+7;
const int N=2e5+10;
typedef long long ll;
typedef pair<int,int> PII;
typedef unsigned long long ull;
inline void read(int &x){
    char t=getchar();
    while(!isdigit(t)) t=getchar();
    for(x=t^48,t=getchar();isdigit(t);t=getchar()) x=x*10+(t^48);
}
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);

	int n,s;
	cin>>n>>s;
	if(n<=s/2){
		cout<<"YES"<<endl;
		for(int i=0;i<n-1;i++){
			cout<<2<<" ";
			s-=2;
		}
		cout<<s<<endl<<1<<endl;
	}
	else
		cout<<"NO"<<endl;
	return 0;
}