牛客网第一场E题 Removal

链接：https://www.nowcoder.com/acm/contest/139/E
来源：牛客网

Bobo has a sequence of integers s1, s2, ..., sn where 1 ≤ si ≤ k.
Find out the number of distinct sequences modulo (109+7) after removing exactly m elements.
输入描述:
The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains three integers n, m and k.
The second line contains n integers s1, s2, ..., sn.
输出描述:
For each test case, print an integer which denotes the result.
示例1
输入
复制
3 2 2
1 2 1
4 2 2
1 2 1 2
输出
复制
2
4
备注:
* 1 ≤ n ≤ 105
* 1 ≤ m ≤ min{n - 1, 10}
* 1 ≤ k ≤ 10
* 1 ≤ si ≤ k
* The sum of n does not exceed 106.

dp [i][j]：表示以i位子结尾删除j个字符的方案数。

开一个nt[i][j] 表示从i开始，字符j的位子

那么我们可以得到一个递推式。

dp[nt[i][p]][nt[i][p]-i-1+j]+=dp[i][j];
dp[i][j]可以转移到下一个位子;
比如 
XXXCYYYC；
那么我们删到xxx和xxxcyyy是一样
所以我们每次找到下一个位子

#include <iostream>
#include<stdio.h>
#include<string.h>
#define ll long long
#define mod 1000000007
using namespace std;
int nt[100005][15];
int a[100005];
int dp[100005][15];//以a[i]结尾的，删了 j个
int main()
{
    int n,m,k;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1; i<=k; i++)
        {
            nt[n][i]=n+1;
        }
        for(int i=n; i>=1; i--)
        {
            for(int j=1; j<=k; j++)
                nt[i-1][j]=nt[i][j];
               nt[i-1][a[i]]=i;
        }
        memset(dp,0,sizeof dp);
        dp[0][0]=1;
         for(int i=0;i<=n;i++)
         {
             for(int j=0;j<=m;j++)
             {
                 for(int p=1;p<=k;p++)
                 {
                     int to=nt[i][p];
                     int shan=to-i-1;
                     if(shan+j<=m)
                     {
                         dp[to][shan+j]+=dp[i][j];
                         dp[to][shan+j]=dp[to][shan+j]%1000000007;
                     }
                 }
             }
         }
         int ans=0;
         for(int i=0;i<=m;i++)
         {
                ans=ans+dp[n-i][m-i];
                ans=ans%1000000007;
         }
          cout<<ans<<endl;

    }


        return 0;
    }