lightoj-1122-Digit Count【DP】

题目链接：点击打开链接

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

题解：又是DP

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int m,n;int a[50];int dp[50][50]; // dp[i][j]=k 表示前 i个数组成 j位数有 k种方法 int main(){int t,text=0;scanf("%d",&t);while(t--){scanf("%d %d",&m,&n);memset(dp,0,sizeof(dp));for(int i=1;i<=m;i++){scanf("%d",&a[i]);dp[i][1]=1;//dp[1][i]=1; 这种方法是 dp[i][j]=k 表示前 j个数组成 i位数有 k种方法  }for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){for(int k=1;k<=m;k++){if(abs(a[j]-a[k])<=2){dp[j][i]+=dp[k][i-1];//dp[i][j]+=dp[i-1][k];}}}}int ans=0;for(int i=1;i<=m;i++)ans+=dp[i][n];//ans+=dp[n][i];printf("Case %d: %d\n",++text,ans);}return 0;}

搜索

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int m,n,ans;int a[50];void find(int x,int cnt){if(cnt==n) // 搜到长度为 n的时候就停止 搜索 {ans++;return ;}for(int i=0;i<m;i++){if(abs(a[i]-x)<=2)find(a[i],cnt+1);}}int main(){int t,text=0;scanf("%d",&t);while(t--){scanf("%d %d",&m,&n);for(int i=0;i<m;i++)scanf("%d",&a[i]);ans=0;for(int i=0;i<m;i++)find(a[i],1);printf("Case %d: %d\n",++text,ans);}return 0;}