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.

<h1>Sample Input</h1> <h1>Output for Sample Input</h1>
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;
}

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