A. Greed
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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).
Examples
Input
Copy
2 3 5 3 6
Output
Copy
YES
Input
Copy
3 6 8 9 6 10 12
Output
Copy
NO
Input
Copy
5 0 0 5 0 0 1 1 8 10 5
Output
Copy
YES
Input
Copy
4 4 1 0 3 5 2 2 3
Output
Copy
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
题目链接：http://codeforces.com/contest/892/problem/A
题意：有n瓶cola，然后每瓶的体积是ai，容量是bi，问是否能用两个 瓶子将所有的cola装完。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
int a[maxn];
int main()
{
int n;
while(~scanf("%d", &n))
{
ll sum = 0;
for(int i = 0; i < n; i++)
{
int x;
cin >> x;
sum += x;
}
for(int i = 0; i < n; i++)
cin >> a[i];
sort(a, a+n);
ll ans = a[n-1] + a[n-2];
//cout << ans << sum << endl;
if(ans >= sum)
puts("YES");
else
puts("NO");
}
return 0;
}