CF 348A(Mafia-判定性问题)

A. Mafiatime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other n?-?1people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the i-th person wants to play a_{i rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?}

Input

The first line contains integer n (3?≤?n?≤?10^{5). The second line contains n space-separated integers a_{1,?a2,?...,?an (1?≤?ai?≤?109) — the i-th number in the list is the number of rounds the i-th person wants to play.}}

Output

In a single line print a single integer — the minimum number of game rounds the friends need to let the i-th person play at least a_irounds.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Sample test(s)input

33 2 2

output

4

input

42 2 2 2

output

3

Note

You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times:http://en.wikipedia.org/wiki/Mafia_(party_game).

这题如果硬解没思路，但是判断一个答案就是很简单。

然后你懂得。。

因为用了int惨遭Hack，各种NC。。。

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (100000+10)long long mul(long long a,long long b){return (a*b)%F;}long long add(long long a,long long b){return (a+b)%F;}long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}typedef long long ll;int n;ll a[MAXN];int main(){cin>>n;For(i,n) cin>>a[i];ll k=0;For(i,n) k=max(k,a[i]);ll t=0;For(i,n) t+=k-a[i];if (t>=k) {cout<<k<<endl;return 0;}else {double p=((double)(k-t))/((double)(n-1));k+=ceil(p);cout<<k<<endl;}return 0;}