CF996B World Cup 思维 第十四 *

World Cup
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Allen wants to enter a fan zone that occupies a round square and has nn entrances.

There already is a queue of aiai people in front of the ii-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.

Allen uses the following strategy to enter the fan zone:

  • Initially he stands in the end of the queue in front of the first entrance.
  • Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).

Determine the entrance through which Allen will finally enter the fan zone.

Input

The first line contains a single integer nn (2n1052≤n≤105) — the number of entrances.

The second line contains nn integers a1,a2,,ana1,a2,…,an (0ai1090≤ai≤109) — the number of people in queues. These numbers do not include Allen.

Output

Print a single integer — the number of entrance that Allen will use.

Examples
input
Copy
4
2 3 2 0
output
Copy
3
input
Copy
2
10 10
output
Copy
1
input
Copy
6
5 2 6 5 7 4
output
Copy
6
Note

In the first example the number of people (not including Allen) changes as follows: [2,3,2,0][1,2,1,0][0,1,0,0][2,3,2,0]→[1,2,1,0]→[0,1,0,0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.

In the second example the number of people (not including Allen) changes as follows:[10,10][9,9][8,8][7,7][6,6][5,5][4,4][3,3][2,2][1,1][0,0][10,10]→[9,9]→[8,8]→[7,7]→[6,6]→[5,5]→[4,4]→[3,3]→[2,2]→[1,1]→[0,0].

In the third example the number of people (not including Allen) changes as follows:[5,2,6,5,7,4][4,1,5,4,6,3][3,0,4,3,5,2][2,0,3,2,4,1][1,0,2,1,3,0][0,0,1,0,2,0]

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
ll a[maxn];
int main(){
    std::ios::sync_with_stdio(false);
    ll n;
    while( cin >> n ) {
        ll maxm = 1e12, sum = n;
        for( ll i = 1; i <= n; i ++ ) {
            cin >> a[i];
            a[i] -= i - 1;
            if( maxm > ( a[i] + n - 1 ) / n ) {
                maxm = ( a[i] + n - 1 ) / n;
                sum = i;
            }
        }
        cout << sum << endl;
    }
    return 0;
}

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转载自www.cnblogs.com/l609929321/p/9225543.html