Codeforces Round #257 (Div. 2) A. Jzzhu and Children

A. Jzzhu and Children

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from1 to n. The i-th child wants to get at least a_i candies.

Jzzhu asks children to line up. Initially, the i-th child stands at thei-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:

1. Give m candies to the first child of the line.
2. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home.
3. Repeat the first two steps while the line is not empty.

Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?

Input

The first line contains two integers n, m(1 ≤ n ≤ 100; 1 ≤ m ≤ 100). The second line containsn integers a₁, a₂, ..., a_n(1 ≤ a_i ≤ 100).

Output

Output a single integer, representing the number of the last child.

Sample test(s)

Input

5 21 3 1 4 2

Output

4

Input

6 41 1 2 2 3 3

Output

6

Note

Let's consider the first sample.

Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.

Child 4 is the last one who goes home.

题意：给n个小朋友发糖，一次给k个，每个人有想要的糖数，如果给了一个小朋友糖后，没达到他想要的糖数，那么他会到最后一个位置，然后等一轮后继续给，直到达到他想要的糖数，给够了就回家，问最后一个回家的小朋友是第几个

思路：直接找给的次数最多的最后一个就行了

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)using namespace std;LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}LL lcm(LL a,LL b){return a/gcd(a,b)*b;}LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}//headint main(){int n,m;while(scanf("%d%d",&n,&m)!=EOF){int M=-1;int ans;for(int i=1;i<=n;i++){int a;scanf("%d",&a);int k=a/m;if(a%m)k++;if(k>=M)M=k,ans=i;}printf("%d\n",ans);}return 0;}