【Codeforces Round 859 (Div. 4)】G2. Subsequence Addition (Hard Version)题解

题目大意
The only difference between the two versions is that in this version, the constraints are higher.

Initially, array $a$ contains just the number $1$. You can perform several operations in order to change the array. In an operation, you can select some subsequence${}^{†}$ of $a$ and add into $a$ an element equal to the sum of all elements of the subsequence.

You are given a final array $c$. Check if $c$ can be obtained from the initial array $a$ by performing some number (possibly 0) of operations on the initial array.

${}^{†}$ A sequence $b$ is a subsequence of a sequence $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly zero, but not all) elements. In other words, select $k$ ($1\le k\le |a|$) distinct indices $i_1,i_2,\dots ,i_k$ and insert anywhere into $a$ a new element with the value equal to $a_{i_1}+a_{i_2}+\cdots +a_{i_k}$.

Input

The first line of the input contains an integer $t$ ($1\le t\le 1000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$) — the number of elements the final array $c$ should have.

The second line of each test case contains $n$ space-separated integers $c_i$ ($1\le c_i\le 2\cdot 10^5$) — the elements of the final array $c$ that should be obtained from the initial array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output

For each test case, output “YES” (without quotes) if such a sequence of operations exists, and “NO” (without quotes) otherwise.

You can output the answer in any case (for example, the strings “yEs”, “yes”, “Yes” and “YES” will be recognized as a positive answer).
Example
input

6
1
1
1
2
5
5 1 3 2 1
5
7 1 5 2 1
3
1 1 1
5
1 1 4 2 1

output

YES
NO
YES
NO
YES
YES

Note

For the first test case, the initial array $a$ is already equal to $[1]$, so the answer is “YES”.

For the second test case, performing any amount of operations will change $a$ to an array of size at least two which doesn’t only have the element $2$, thus obtaining the array $[2]$ is impossible and the answer is “NO”.

For the third test case, we can perform the following operations in order to obtain the final given array $c$:

• Initially, $a=[1]$.
• By choosing the subsequence $[1]$, and inserting $1$ in the array, $a$ changes to $[1,1]$.
• By choosing the subsequence $[1,1]$, and inserting $1+1=2$ in the middle of the array, $a$ changes to $[1,2,1]$.
• By choosing the subsequence $[1,2]$, and inserting $1+2=3$ after the first $1$ of the array, $a$ changes to $[1,3,2,1]$.
• By choosing the subsequence $[1,3,1]$ and inserting $1+3+1=5$ at the beginning of the array, $a$ changes to $[5,1,3,2,1]$ (which is the array we needed to obtain).

题目大意

题解

一
<<<<<思路

我们首先说正统的题解，使用数学结论的方法，我们知道这个数从1开始，特判1 ，后面的数都是前面的数的一部分的和，我们只知道他至少有两个1 和 一个2 ，1，2，3，4可以表示，当我们取最大的时候最不容易表示8 。。。数学归纳最后可知这个数组中的最大值只要不超过和就能一直成立

<<<<<<代码

    #include <iostream>
    #include <algorithm> 
    using namespace std ;
     
    const int N = 200010 ;
     
    void work() 
    {
    
    	int q[N] ;
    	int n ;
    	cin >> n ;
    	for(int i = 0 ; i < n ; i ++ )
    		cin >> q[i] ;
    	sort (q, q + n) ;
    	if(q[0] != 1) 
    	{
    
    
    		cout << "NO" << endl ;
    		return ;
    	}
    	long long  sum = 1 ;
    	for(int i = 1 ; i < n ; i ++  )
    	{
    
    
     
    		if(sum < q[i])
    		{
    
    
    			
    			cout <<"NO" << endl ;
    			return ;
    		}
    		sum = sum + q[i] ;
    	}
    	cout << "YES" <<endl ;
    	return ;
    	
     
    }
     
    int main ()
    {
    
    
    	int T ;
    	cin >> T ;
    	while ( T -- )
    	{
    
    
    		work() ;
    	}
    	return 0 ;
    }

<<<<< 以上思路为题目的正解，但是这里提供一种其他的思路也能解决类似问题，但是思路不一样，时间更长。
二
本题解设计到数据结构 bitset,详情请看。但这个并不是正确的答案，在G2 会TLE ,但是了解这种数据结构，在没有归纳出来规律的时候可以直接A G1

#include <iostream>
#include <bitset>
#include <algorithm>
using namespace std ;
const int N = 6000 ; 

bitset <N> b ;
int q[N] ;
void work() 
{
    
    
	int n ;
	cin >> n ;
	for(int i = 0 ; i < n ;  i++ )
	{
    
    
		cin >> q[i] ;
	}	
	sort(q , q + n ) ;
	if( q[0] != 1 )
	{
    
    
		cout << "NO" << endl ;
		return ;
	}
	int sum = 1 ;
	for(int i = 1 ; i < n ; i ++ )
	{
    
    
		if(sum < q[i])
		{
    
    
			cout << "NO" << endl ;
			return ;
		}
		sum += q[i] ;
	}
	cout << "YES" << endl ;
	return ;

}

int main () 
{
    
    
	int n ;
	cin >>  n ;
	while ( n -- )
	{
    
    
		work() ;
	}
	return 0 ;
}