lightoj-1122-Digit Count【DP】

题目链接：​​点击打开链接​​

1122 - Digit Count

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.

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;
}