Table Tennis
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the
next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 ≤ n ≤ 500, 2 ≤ k ≤ 1012) — the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) — powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer — power of the winner.
Example
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
题意：如果a>b,a就赢了当a赢了k次，就输出a;
思路：记录每个赢得人赢得次数，谁赢了k次就输出
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;
int a[1000];
int b[1000];
int main()
{
int n;
long long k;
scanf("%d%lld",&n,&k);
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(int i=0; i<n; i++)
{
scanf("%d",&a[i]);
}
int x=0;
for(int i=1; i<n; i++)
{
if(a[x]>a[i]&&a[i]!=-1)
{
a[i]=-1;
b[x]++;
}
else if(a[x]<a[i]&&a[i]!=-1)
{
x=i;
b[x]++;
}
if(b[x]==k)
{
printf("%d
",a[x]);
return 0;
}
//cout<<x<<endl;
}
printf("%d
",a[x]);
}