CodeForce -1355 Game With Array 【思维 + 构造】题解

忘是亡心i 2021-07-24 20:02 359阅读 0赞

### 目录 ###

*   *   *  1.题目
         *  2.思路
         *  3.代码

### 1.题目 ###

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

### 2.思路 ###

**题意:**  
给出你数组长度n和数组的所有元素和s，让你构造出一个数组，使得该数组的任意子序列和都不为k或s-k。  
**思路:**  
题目的关键之处在于**该数组的任意子序列和都不为k或s-k**。k的取值范围为1<=k<=s。  
如果我们构造这样一个数组 a\[1\]=1,a\[2\]=1,a\[3\]=1…直到a\[k-1\]=1.  
也就是说该数组的前k-1项都是1，那么前k-1项的任意子序列和都  
小于k，同样如果我们可以使得该数组的后n-(k-1）项的最小值为k+1，那么该数组无论怎样取子序列，其子序列和要么比k小，要么比k大，反正不会等于k或s-k。问题来了，k取多少合适呢？k取n最合适，k取n时，1有n-1项，比k大的只有一项，其值为s-(n-1)，最为简单。  
构造的数组如下:  
a\[1\]=1,a\[2\]=1,a\[3\]=1…a\[n-1\]=1,a\[n\]=s-(n-1）  
如果s-(n-1)<=k（最后一项的值小于等于k），构造失败。

### 3.代码 ###

#include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    int main()
    { 
    	int n,s,k;
    	cin>>n>>s;
    	k=n;
    	if(s-n+1<=k)  puts("NO");
    	else 
    	{ 
    		puts("YES");
    		for(int i=1;i<=n-1;i++)
    		{ 
    			cout<<'1'<<' ';
    		}
    		cout<<s-n+1<<endl;
    		cout<<k<<endl;
    	}
    	return 0;
    }