Atcoder AtCoder Regular Contest 079 E - Decrease (Judge ver.)

E - Decrease (Judge ver.)

Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N−1 or smaller. (The operation is the same as the one in Problem D.)

Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.

It can be proved that the largest element in the sequence becomes N−1 or smaller after a finite number of operations.

You are given the sequence ai. Find the number of times we will perform the above operation.

Constraints

2≤N≤50
0≤ai≤1016+1000

Input

Input is given from Standard Input in the following format:

N
a1 a2 ... aN

Output

Print the number of times the operation will be performed.

Sample Input 1

Copy

4
3 3 3 3

Sample Output 1

Copy

0

Sample Input 2

Copy

3
1 0 3

Sample Output 2

Copy

1

Sample Input 3

Copy

2
2 2

Sample Output 3

Copy

2

Sample Input 4

Copy

7
27 0 0 0 0 0 0

Sample Output 4

Copy

3

Sample Input 5

Copy

101000 193 256 777 0 1 1192 1234567891011 48 425

Sample Output 5

Copy

1234567894848

模拟一下就可以了

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll n,k,x,m;
priority_queue<ll>q;
int main()
{
ios();
scanf("%lld",&n);
for(int i=1;i<=n;i++)
{
scanf("%lld",&x);
q.push(x);
}
k=0;
while(q.top()+k>=n)
{
ll ans=q.top()+k;
q.pop();
m=ans/n;
ans-=m*n;
k+=m;
q.push(ans-k);
}
printf("%lld\n",k);
return 0;
}