You are playing a variation of game 2048. Initially you have a multiset ss of nn integers. Every integer in this multiset is a power of two.
You may perform any number (possibly, zero) operations with this multiset.
During each operation you choose two equal integers from ss, remove them from ss and insert the number equal to their sum into ss.
For example, if s={1,2,1,1,4,2,2}s={1,2,1,1,4,2,2} and you choose integers 22 and 22, then the multiset becomes {1,1,1,4,4,2}{1,1,1,4,4,2}.
You win if the number 20482048 belongs to your multiset. For example, if s={1024,512,512,4}s={1024,512,512,4} you can win as follows: choose 512512 and 512512, your multiset turns into {1024,1024,4}{1024,1024,4}. Then choose 10241024 and 10241024, your multiset turns into {2048,4}{2048,4} and you win.
You have to determine if you can win this game.
You have to answer qq independent queries.
The first line contains one integer qq (1≤q≤1001≤q≤100) – the number of queries.
The first line of each query contains one integer nn (1≤n≤1001≤n≤100) — the number of elements in multiset.
The second line of each query contains nn integers s1,s2,…,sns1,s2,…,sn (1≤si≤2291≤si≤229) — the description of the multiset. It is guaranteed that all elements of the multiset are powers of two.
For each query print YES if it is possible to obtain the number 20482048 in your multiset, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
6
4
1024 512 64 512
1
2048
3
64 512 2
2
4096 4
7
2048 2 2048 2048 2048 2048 2048
2
2048 4096
YES
YES
NO
NO
YES
YES
In the first query you can win as follows: choose 512512 and 512512, and ss turns into {1024,64,1024}{1024,64,1024}. Then choose 10241024 and 10241024, and ssturns into {2048,64}{2048,64} and you win.
In the second query ss contains 20482048 initially.
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
typedef long long lli;
#define pql priority_queue<ll>
#define pq priority_queue<int>
#define v vector<int>
#define vl vector<ll>
//线段树
#define lson rt<<1, l, m
#define rson rt<<1|1, m+1, r
#define gcd __gcd
#define mem(s,t) memset(s,t,sizeof(s))
#define debug(a,n) for(int i=0;i<n;i++) cout<<" "<<a[i];
#define Debug(a,n,m) for(int i=0;i<n;i++) {for(int j=0;j<m;j++) cout<<a[i][j]<<" "; cout<<endl; }
#define rep(j,k) for (int i = j; i <= k; i++)
#define per(i,k) for (int i = i; i >= k; i--)
#define test cout<<"+++"<<endl;
#define input(a,k) for (int i = 1; i <= (int)(k); i++) {scanf("%d",&a[i]) ; }
#define INPUT(a,n,m) for(int i=0;i<n;i++) {for(int j=0;j<m;j++) cin>>a[i][j] ; }
#define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ok return 0;
#define oi(x) cout<<x<<endl;
#define os(str) cout<<string(str)<<endl;
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
lli Fibonacci(lli n) { return (1/sqrt(5))*( pow( (1+sqrt(5))/2,n )-pow( (1-sqrt(5))/2 ,n) ); }
lli FibonacciSum(lli n) { return 2*Fibonacci(n)+Fibonacci(n-1)-1; }
using namespace std;
const int mxn = 2e5+5;
int t,n,m,flag,a[mxn],k;
map<lli,int>mp;
int main()
{
TLE;
cin>>t;
lli sum;
while(t--)
{
mem(a,0);
sum=0; flag = 0;
cin>>n;
rep(1,n)
{
cin>>k;
if(k==2048) flag=true;
k %= 2048;
a[k]++;
sum+=a[k]*k;
}
if(flag) {YES;}
else if(sum<2048) {NO;}
else
{
k=1;
for(int i=1;i<=11;i++)
{
a[k<<1] += a[k]/2;
k<<=1;
}
if(a[2048]) {YES;}
else {NO;}
}
}
ok;
}