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).

In the first sample, there are already 2 cans, so the answer is "YES".

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+5;
ll l[N],ma[N];
int main()
{ios::sync_with_stdio(false);cin.tie(0);ll n;cin>>n;memset(ma,0,sizeof(ma));for(ll i=1;i<=n;i++){cin>>l[i];ll x=max(1LL,i-l[i]);ma[x]=max(i-x,ma[x]);}ll ans=0,f=0;for(ll i=1;i<=n;i++){f=max(f,ma[i]);if(f==0)ans++;f--;}cout<<ans;return 0;
}

You have an array a with length n, you can perform operations. Each operation is like this: choose two adjacentelements from a, say x and y, and replace one of them with gcd(x, y), where gcd denotes the greatest common divisor.

What is the minimum number of operations you need to make all of the elements equal to 1?

The first line of the input contains one integer n (1 ≤ n ≤ 2000) — the number of elements in the array.

The second line contains n space separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the array.

Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1.

In the first sample you can turn all numbers to 1 using the following 5 moves:

• [2, 2, 3, 4, 6].
• [2, 1, 3, 4, 6]
• [2, 1, 3, 1, 6]
• [2, 1, 1, 1, 6]
• [1, 1, 1, 1, 6]
• [1, 1, 1, 1, 1]

We can prove that in this case it is not possible to make all numbers one using less than 5 moves.

这个有1的话就得直接输出啊，n-f

#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
const int N=2005;
int a[N],b[N];
int main()
{ios::sync_with_stdio(false);cin.tie(0);int n;cin>>n;int f=0;for(int i=1; i<=n; i++){cin>>a[i];if(a[i]==1)f++;}if(f)printf("%d",n-f);else{f=1;for(;;){if(f>888*n){printf("-1");return 0;}for(int i=f+1; i<=n; i++){b[i]=__gcd(a[i-1],a[i]);if(b[i]==1){cout<<f+n-1;return 0;}}f++;for(int i=1; i<=n; i++)a[i]=b[i];}}return 0;
}

21 2

2 1 

41000 100 10 1

100 1 1000 10

#include<bits/stdc++.h>
using namespace std;
const int N=200;
pair<int,int>a[N];
int ans[N];
int main()
{int n;scanf("%d",&n);for(int i=0; i<n; i++)scanf("%d",&a[i].first),a[i].second=i;sort(a,a+n);for(int i=0; i<n; i++){int to=a[i].second;if(i==0)ans[to]=a[n-1].first;elseans[to]=a[i-1].first;}for(int i=0; i<n; i++){printf("%d ",ans[i]);}return 0;
}