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".
题意:有 n 瓶可乐 , 然后每瓶可乐都有剩余的可乐 和 不同的容量 , 问能不能把所有剩余的可乐倒在 两个可乐瓶子里。
思路:累加可乐剩余的体积 , 找出 最大的两个可乐的 容量和 , 比较
#include <iostream> #include <algorithm> using namespace std ; #define maxn 101000 #define LL long long int main(){ int n ; LL num[maxn] ; LL sum[maxn] ; while(~scanf("%d" , &n)){ LL summ = 0 ; for(int i=1 ; i<= n ; i++){ scanf("%lld" , &num[i]) ; summ += num[i] ; } for(int i=1 ; i<=n ; i++){ scanf("%lld" , &sum[i]) ; } sort(sum+1 , sum+1+n ) ; if(summ <= sum[n] + sum[n-1]) { printf("YES ") ; } else { printf("NO ") ; } } return 0 ; }