A. Greed
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can’s capacity bi (ai  ≤  bi).

Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!

Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.

The second line contains n space-separated integers a1, a2, …, an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.

The third line contains n space-separated integers that b1, b2, …, bn (ai ≤ bi ≤ 109) — capacities of the cans.

Output
Print “YES” (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print “NO” (without quotes).

You can print each letter in any case (upper or lower).

My Answer Code:

/*
	Author:Albert Tesla Wizard
	Time:2020/10/29 14:37
*/
#include<bits/stdc++.h>
using namespace std;
using ull=unsigned long long;
struct can
{
    
    
    ull volume;
    ull capacity;
};
bool cmp(can a,can b)
{
    
    
    return a.capacity<=b.capacity;
}
int main()
{
    
    
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    ull sum=0;
    cin>>n;
    vector<can>a(n);
    for(int i=0;i<n;i++){
    
    cin>>a[i].volume;}
    for(int i=0;i<n;i++)cin>>a[i].capacity;
    if(n==2){
    
    cout<<"YES"<<endl;return 0;}
    for(int i=0;i<n;i++){
    
    sum+=a[i].volume;if(sum>2000000000){
    
    cout<<"NO"<<endl;return 0;}}
    sort(a.begin(),a.end(),cmp);
    if(a[n-2].capacity+a[n-1].capacity>=sum){
    
    cout<<"YES"<<endl;}
    else cout<<"NO"<<endl;
    return 0;
}