A. Greed(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

题解：

注意开long long
计算出总的volume与前两大的capacity进行比较即可

代码：

#include <bits/stdc++.h>using namespace std;const int maxn = 1e5+100;typedef long long LL;struct Cola{    LL volume;    LL capacity;};Cola cola[maxn];bool cmp(Cola a,Cola b){    return a.capacity>b.capacity;}int main(){    int n;    cin>>n;    LL sumV=0;    for(int i=0;i<n;i++)    {      cin>>cola[i].volume;      sumV+=cola[i].volume;    }    for(int i=0;i<n;i++)     cin>>cola[i].capacity;    sort(cola,cola+n,cmp);    LL sumC = cola[0].capacity+cola[1].capacity;    if(sumV<=sumC)    {        cout<<"YES"<<endl;    }else    {        cout<<"NO"<<endl;    }    return 0;}