Codeforces 369D Valera and Fools【思维+dp】

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本文介绍了一个基于概率的游戏生存模拟算法，通过分析每个玩家的射击命中率，在限定的轮次内预测可能的生存状态组合。

One fine morning, n fools lined up in a row. After that, they numbered each other with numbers from 1 to n, inclusive. Each fool got a unique number. The fools decided not to change their numbers before the end of the fun.

Every fool has exactly k bullets and a pistol. In addition, the fool number i has probability of p_i (in percent) that he kills the fool he shoots at.

The fools decided to have several rounds of the fun. Each round of the fun looks like this: each currently living fool shoots at another living fool with the smallest number (a fool is not stupid enough to shoot at himself). All shots of the round are perfomed at one time (simultaneously). If there is exactly one living fool, he does not shoot.

Let's define a situation as the set of numbers of all the living fools at the some time. We say that a situation is possible if for some integer number j (0 ≤ j ≤ k) there is a nonzero probability that after j rounds of the fun this situation will occur.

Valera knows numbers p₁, p₂, ..., p_n and k. Help Valera determine the number of distinct possible situations.

Input

The first line contains two integers n, k (1 ≤ n, k ≤ 3000) — the initial number of fools and the number of bullets for each fool.

The second line contains n integers p₁, p₂, ..., p_n (0 ≤ p_i ≤ 100) — the given probabilities (in percent).

Output

Print a single number — the answer to the problem.

Examples

Input

3 3
50 50 50

Output

Input

1 1
100

Output

Input

2 1
100 100

Output

Input

3 3
0 0 0

Output

Note

In the first sample, any situation is possible, except for situation {1, 2}.

In the second sample there is exactly one fool, so he does not make shots.

In the third sample the possible situations are {1, 2} (after zero rounds) and the "empty" situation {} (after one round).

In the fourth sample, the only possible situation is {1, 2, 3}.

题目大意：

给你N个人，以及要进行K轮游戏，每个人都有一把枪，每个人一开始都将抢指向在场上编号最小的那个人身上，编号最小的那个人将枪口指向编号次小的那个人身上。

现在已知每个人打中的概率，问K轮内，会有几种存活情况出现。

思路：

特征类Dp，寻找问题特征，问题的焦点在于编号最小和次小的两个人身上，那么我们设定Dp【i】【j】表示此时场上剩下的人编号最小的人是i，次小的人是j的情况发生的最早轮数.

那么考虑对于这两个人的四种情况：

①i死了，j没有死。那么需要满足的条件有：从j编号开始到n编号，打死i的最大概率应>0.同时pi<100.

Dp【j】【j+1】=min（Dp【j】【j+1】，dp【i】【j】+1）；

②i死了，j也死了。那么需要满足的条件有：从j编号开始到n编号，打死i的最大概率应>0.同时pi>0.

Dp【j+1】【j+2】=min（Dp【j+1】【j+2】，dp【i】【j】+1）；

③i没死，j死了。那么需要满足的条件有：从j编号开始到n编号，打死i的最大概率应<100.同时pi>0.

Dp【i】【j+1】=min（Dp【i】【j+1】，dp【i】【j】+1）；

④i没死，j也没死。那么没有继续转移下去的意义。

那么ans=sum（dp【i】【j】<=k）；

注意初始化以及最终剩余1个人、两个人的情况。

Ac代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int p[5000];
int dp[3005][3005];
int back[3005];
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        memset(back,0,sizeof(back));
        memset(p,0,sizeof(p));
        for(int i=1;i<=3003;i++)
        {
            for(int j=1;j<=3003;j++)
            {
                dp[i][j]=0x3f3f3f3f;
            }
        }
        for(int i=1;i<=n;i++)scanf("%d",&p[i]);
        for(int i=n;i>=1;i--)
        {
            if(i==n)back[i]=p[i];
            else back[i]=max(back[i+1],p[i]);
        }
        dp[1][2]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                if(back[j]>0&&p[i]<100)dp[j][j+1]=min(dp[j][j+1],dp[i][j]+1);
                if(back[j]>0&&p[i]>0)dp[j+1][j+2]=min(dp[j+1][j+2],dp[i][j]+1);
                if(back[j]<100&&p[i]>0)dp[i][j+1]=min(dp[i][j+1],dp[i][j]+1);
            }
        }
        int output=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n+1;j++)
            {
                if(dp[i][j]<=k)output++;
            }
        }
        if(dp[n+1][n+2]<=k)output++;
        printf("%d\n",output);
    }
}