2 seconds
256 megabytes
standard input
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!
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.
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).
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
In the first sample, there are already 2 cans, so the answer is "YES".
分析:签到题,排个序看容量最大的杯子能否装下.
#include <cstdio>
#include <cmath>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; long long n, a[], b[], sum; int main()
{
cin >> n;
for (int i = ; i <= n; i++)
{
long long t;
cin >> t;
sum += t;
}
for (int i = ; i <= n; i++)
cin >> b[i];
sort(b + , b + + n);
if (sum <= b[n] + b[n - ])
puts("YES");
else
puts("NO"); return ;
}