Light oj 1122 Digit Count（简单dp）

1122 - Digit Count

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Given a set of digits S, and an integer n, you have to find how many n-digit integers are there, which contain digits that belong to S and the difference between any two adjacent digits
is not more than two.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case contains two integers, m (1 ≤ m < 10) and n (1 ≤ n ≤ 10). The next line will contain m integers (from 1 to 9) separated by spaces. These integers form the set S as
described above. These integers will be distinct and given in ascending order.

Output

For each case, print the case number and the number of valid n-digit integers in a single line.

Sample Input	Output for Sample Input
3 3 2 1 3 6 3 2 1 2 3 3 3 1 4 6	Case 1: 5 Case 2: 9 Case 3: 9

Note

For the first case the valid integers are

11

13

31

33

66

PROBLEM SETTER: JANE ALAM JAN

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-12

typedef __int64 ll;

using namespace std;
#define INF 0x3f3f3f3f
#define N 15

int vis
;
int dp

;

int n,m;

int main()
{
    int i,j,t,ca=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&n);
        memset(vis,0,sizeof(vis));
        memset(dp,0,sizeof(dp));
        int x;
        while(m--)
        {
            scanf("%d",&x);
            vis[x]=1;
        }
        for(i=1;i<=9;i++)
            if(vis[i]) dp[1][i]=1;

       for(int i=2;i<=n;i++)
          for(int j=1;j<=9;j++)
            for(int k=1;k<=9;k++)
               if(vis[j]&&abs(j-k)<=2)
                  dp[i][j]+=dp[i-1][k];

       int ans=0;
       for(int i=1;i<=9;i++)
           ans+=dp
[i];

       printf("Case %d: %d\n",++ca,ans);
    }
   return 0;
}