A - Greed CodeForces - 892A（水题）

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).

Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is “YES”.
思路：大水题，模拟即可。只不过又忘记开long long了结果卡在第五组数据。

#include <bits/stdc++.h>//第五组数据特别大，n是1e5,num_a[i]都是1e9，不知道为什么就是过不了这组
#define ll long long//额，找到问题了，又忘记开long long 了。。。

using namespace std;
const int maxn = 1e5 + 5;

int main()
{
    ll n;
    ll num_a[maxn];
    ll num_b[maxn];
    scanf("%d",&n);
    ll sum = 0;
    for(int i = 0;i < n;i++)
    {
        scanf("%lld",&num_a[i]);
        sum += num_a[i];
    }
    for(int i = 0;i < n;i++)
    {
        scanf("%lld",&num_b[i]);
    }
    sort(num_b,num_b + n);
    int Max1 = num_b[n - 1];
    int Max2 = num_b[n - 2];
    if(sum <= Max1 + Max2)
    {
        printf("YES\n");
    }
    else printf("NO\n");
    return 0;
}