CodeForces - 996B

标签：ssi   iostream   eterm   pie   line   note   closed   single   deque   

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 (2≤n≤1052≤n≤105) — the number of entrances.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1090≤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

4
2 3 2 0

3

2
10 10

1

6
5 2 6 5 7 4

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][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].

　　第一秒的时候这个人站在第一个数上，这个人每秒钟向前移动一步，每移动一步区间中的所有的数都会减一，直到他恰好走到一个数减为零的地方，输出该位置的序号。

　　显而易见。只是一个循环，周期已知，步长已知，故而可推算出每个数在第几圈的时候减为零 nk*(i-1)==Ai (n是序列的长的，k是最少需要的圈数，i是位置的序列，Ai是第i个数的值) ，k最小的那个位置即为答案。

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<string>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<stack>
 9 #include<deque>
10 #include<map>
11 #include<iostream>
12 using namespace std;
13 typedef long long  LL;
14 const double pi=acos(-1.0);
15 const double e=exp(1);
16 const int N = 10;
17 
18 LL con[100009];
19 int main()
20 {
21     LL i,p,j,n;
22     LL mid;
23     scanf("%lld",&n);
24     for(i=0;i<n;i++)
25     {
26         scanf("%lld",&con[i]);
27         mid=ceil((double)(con[i]-i)/n);
28         con[i]=mid;
29     }
30     LL head=999999999999,spot=1;
31     for(i=0;i<n;i++)
32     {
33         if(con[i]<head)
34         {
35             head=con[i];
36             spot=i+1;
37         }
38     }
39     printf("%lld\n",spot);
40     return 0;
41 }

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CodeForces - 996B

标签：ssi   iostream   eterm   pie   line   note   closed   single   deque   

原文地址：https://www.cnblogs.com/daybreaking/p/9694777.html