不要62（数位dp模板题）

 

 AC_Code:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <string>
 5 #include <cmath>
 6 #include <queue>
 7 #include <stack>
 8 #include <vector>
 9 #include <set>
10 #include <map>
11 #include <algorithm>
12 using namespace std;
13 typedef long long ll;
14 const int maxn=10;
15 
16 int dp[maxn][maxn][2];
17 int a[maxn];
18 int len;
19 
20 int dfs(int pos, int pre, int status, bool lead, bool limit){
21     if( pos<1 ) return status;
22     if((!limit)&&(!lead)&&dp[pos][pre][status]!=-1) return dp[pos][pre][status];
23     int up=limit?a[pos]:9;
24     int ans=0;
25     for(int i=0;i<=up;i++){
26         if( pre==6&&i==2 ) continue;
27         else if( i==4 ) continue;
28         ans += dfs(pos-1,i,1,(lead&&(!i)),limit&&(i==up));
29     }
30     return ((!limit)&&(!lead))?dp[pos][pre][status]=ans:ans;
31 }
32 
33 int part(int x){
34     len=0;
35     while(x) a[++len]=x%10,x/=10;
36     memset(dp,-1,sizeof(dp));
37     return dfs(len,-1,1,1,1);
38 }
39 
40 int main()
41 {
42     int n,m;
43     while(~scanf("%d%d",&n,&m)){
44         if( n==0&&m==0 ) break;
45         else if( n==0 ) printf("%d\n",part(m)-part(n)+1);
46         else printf("%d\n",(part(m)-part(n-1)));
47     }
48     return 0;
49 }

 

来源：https://www.cnblogs.com/wsy107316/p/12250028.html