【LightOJ 1122 + dp】

Description
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
3
3 2
1 3 6
3 2
1 2 3
3 3
1 4 6
Sample Output
Case 1: 5
Case 2: 9
Case 3: 9
Hint
For the first case the valid integers are

11
13
31
33
66
Problem Setter: Jane Alam Jan
Developed and Maintained by
JANE ALAM JAN
Copyright © 2012
LightOJ, Jane Alam Jan

一道裸 dp 题 ： dp[i][j] 表示选好前 i - 1个数 第 i 个数以 j 结尾的所有可能结果；

AC 代码 ：

#include<cstdio>#include<cmath>#include<cstring>using namespace std;int pa[12],dp[12][12];int main(){    int T,N,M,i,j,kl,nl = 0;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&M,&N);        for(i = 1 ; i <= M ; i++)            scanf("%d",&pa[i]);            memset(dp,0,sizeof(dp));        for(i = 1 ; i <= M; i++)            dp[1][i] = 1;        for(kl = 2 ; kl <= N ; kl++)            for(i = 1 ; i <= M; i++)                for(j = 1 ; j <= M ; j++)                {                    if(abs(pa[j] - pa[i])<= 2)                      dp[kl][i] += dp[kl - 1][j];                }        int ans = 0;        for(i = 1 ; i <= M ; i++)            ans += dp[N][i];        printf("Case %d: %d\n",++nl,ans);    }    return 0;}