标签：tput index ide class scribe nta ecif span using
题目链接：https://cn.vjudge.net/problem/CodeForces-892A
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).
Example
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
贼JR的水，根本不需要题解……
只是想贴一个记录一下O(n)得到数组内最大和次大的for循环代码，免得以后什么时候脑抽忘记了下不出来僵掉了……
#include<bits/stdc++.h> using namespace std; typedef long long LL; int n; int main() { scanf("%d",&n); LL sum=0; for(int i=1,tmp;i<=n;i++) { scanf("%d",&tmp); sum+=tmp; } LL max1=-1,max2=-2; for(int i=1,tmp;i<=n;i++) { scanf("%d",&tmp); if(max2<=max1 && max1<=tmp) max2=max1,max1=tmp; else if(max2<tmp && tmp<=max1) max2=tmp; } if(max1+max2 >= sum) cout<<"YES"<<endl; else cout<<"NO"<<endl; }
codeforces 892A - Greed - [超级大水题][O(n)数组最大和次大]
标签：tput index ide class scribe nta ecif span using
原文地址：http://www.cnblogs.com/dilthey/p/7875359.html
