Game With Array CodeForces - 1355D（构造）

Petya and Vasya are competing with each other in a new interesting game as they always do.

At the beginning of the game Petya has to come up with an array of N positive integers. Sum of all elements in his array should be equal to S. Then Petya has to select an integer K such that 0≤K≤S.

In order to win, Vasya has to find a non-empty subarray in Petya’s array such that the sum of all selected elements equals to either K or S−K. Otherwise Vasya loses.

You are given integers N and S. You should determine if Petya can win, considering Vasya plays optimally. If Petya can win, help him to do that.

Input
The first line contains two integers N and S (1≤N≤S≤106) — the required length of the array and the required sum of its elements.

Output
If Petya can win, print “YES” (without quotes) in the first line. Then print Petya’s array in the second line. The array should contain N positive integers with sum equal to S. In the third line print K. If there are many correct answers, you can print any of them.

If Petya can’t win, print “NO” (without quotes).

You can print each letter in any register (lowercase or uppercase).

Examples
Input
1 4
Output
YES
4
2
Input
3 4
Output
NO
Input
3 8
Output
YES
2 1 5
4
思路：这一场C和D放反了。。
根据样例，我们可以发现如果可以构造成功的，那么k取任意值都是可以的；如果不能构造成功，那么k取任意值都是不可以的。换句话讲，也就是只要证明了一个k不符合，那么就不可能符合了。我们取一种最简单的情况，k=1的时候，也就是不能出现1和s-1的子序列。
如果s是奇数的话，那么s-1就是偶数，我们令前n-1个都是2，那么最后一个一定是奇数，那么不能出现s-1的情况；s为偶数也是一样的。
那么不能构造的情况就是s<2*n了。
代码如下：

#include
#define ll long long
using namespace std;

int n,s;

int main()
{
	while(~scanf("%d%d",&n,&s))
	{
		if(2*n>s) cout<<"NO"<<endl;
		else
		{
			cout<<"YES"<<endl;
			for(int i=1;i<n;i++) cout<<2<<" ";
			cout<<s-2*(n-1)<<endl;
			cout<<1<<endl;
		}
	}
	return 0;
}

努力加油a啊，(^o)/~