| 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
題意：給你一些數字1~9不重複，這些數可以無限使用，問你可以構成多少種給n位數，并且滿足相鄰的兩位數內插補點不超過2。 思路：dp[i][j]表示第i位為j的方案數，最後累加dp[n][j]就可以的了，這個dp比較簡單。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
using namespace std;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f, N = 5e4 + 10, Mod = 1000000007;
typedef long long ll;
struct node
{
int x, y;
} p[N];
int c[20][20];
ll dp[20][20];
int main()
{
int t, n, m, x;
cin>>t;
for(int cas = 1; cas<=t; cas++)
{
scanf("%d%d", &m, &n);
int vis[10];
memset(vis, 0, sizeof(vis));
memset(dp, 0, sizeof(dp));
for(int i = 1; i<=m; i++)
scanf("%d", &x), vis[x] = 1;
for(int i = 1; i<=9; i++)
if(vis[i]) dp[1][i] = 1LL;
for(int i = 2; i<=n; i++)
for(int j = 1; j<=9; j++)
for(int k = 1; k<=9; k++)
{
if(dp[i-1][k] && vis[j] && abs(k - j)<=2)
{
dp[i][j] += dp[i-1][k];
}
}
ll ans = 0;
for(int i = 1; i<=9; i++)
ans+=dp[n][i];
printf("Case %d: %lld\n", cas, ans);
}
return 0;
}