lightoj 1122 - Digit Count(dp) - dblank

# lightoj 1122 - Digit Count(dp)

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 Time Limit: 2 second(s) Memory Limit: 32 MB

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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.
<h1>Input</h1>
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.
<h1>Output</h1>
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

<h1>Note</h1>
For the first case the valid integers are

11

13

31

33

66
<div>题意：给你一些数字1~9不重复，这些数可以无限使用，问你可以构成多少种给n位数，并且满足相邻的两位数差值不超过2。</div>
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<div>思路：dpi表示第i位为j的方案数，最后累加dpn就可以的了，这个dp比较简单。</div>

#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;
}