題目鏈接：https://cn.vjudge.net/problem/CodeForces-892A

Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola a_i and can‘s capacity b_i (a_i ?≤? b_i).

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

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

The third line contains n space-separated integers that b₁,?b₂,?...,?b_n (a_i?≤?b_i?≤?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).

Example

2
3 5
3 6

YES

3
6 8 9
6 10 12

NO

5
0 0 5 0 0
1 1 8 10 5

YES

4
4 1 0 3
5 2 2 3

YES

賊JR的水，根本不需要題解……

只是想貼一個記錄一下O(n)得到數組內最大和次大的for循環代碼，免得以後什麽時候腦抽忘記了下不出來僵掉了……

#include<bits/stdc++.h>
using
 
 namespace std;
typedef long long LL;
int n;
int main()
{
    scanf("%d",&n);
    LL sum=0;
    for(int i=1,tmp;i<=n;i++)
    {
        scanf("%d",&tmp);
        sum+=tmp;
    }
    LL max1=-1,max2=-2;
    for(int i=1,tmp;i<=n;i++)
    {
        scanf("%d",&tmp);
        if(max2<=max1 && max1<=tmp) max2=max1,max1=tmp;
        else if(max2<tmp && tmp<=max1) max2=tmp;
    }
    if(max1+max2 >= sum) cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
}

codeforces 892A - Greed - [超級大水題][O(n)數組最大和次大]