HDU 4389 X mod f(x) 디지털 dp

M - X mod f(x)
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
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Status 
Practice 
HDU 4389
Appoint description:  System Crawler  (2016-05-06)
Description

Here is a function f(x):
　　　int f ( int x ) {
　　　    if ( x == 0 ) return 0;
　　　    return f ( x / 10 ) + x % 10;
　　　}

　　　Now, you want to know, in a given interval [A, B] (1 <= A <= B <= 10 
9), how many integer x that mod f(x) equal to 0.
 
Input
　　　The first line has an integer T (1 <= T <= 50), indicate the number of test cases. 
　　　Each test case has two integers A, B. 
 
Output
　　　For each test case, output only one line containing the case number and an integer indicated the number of x. 
 
Sample Input

      
      
      
      

       
       
       
       2
1 10
11 20 
      
      
      
      

 
Sample Output

      
      
      
      

       
       
       
       Case 1: 10

       
       
       
       
Case 2: 3 
       
       
       
       
f(x)   1-81      f(x)         
       
       
       
       
ACcode：
       
       
       
       
       
       
       
       
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int dp[10][81][81][81];
int data[10];
int cnt=1;
int dfs(int len,int sum,int mod,int res,int limit){
    if(!len)return sum==mod&&res==0;
    if(!limit&&dp[len][sum][mod][res]!=-1)return dp[len][sum][mod][res];
    int ed=limit?data[len]:9;
    int ans=0;
    for(int i=0;i<=ed;++i)
        ans+=dfs(len-1,sum+i,mod,(res*10+i)%mod,limit&&i==ed);
    return limit?ans:dp[len][sum][mod][res]=ans;
}
int fun(int a){
    int len=0;
    while(a){
        data[++len]=a%10;
        a/=10;
    }
  int ans=0;
  for(int i=1;i<=81;++i)
    ans+=dfs(len,0,i,0,1);
  return ans;
}
void doit(){
    int a,b;
    scanf("%d%d",&a,&b);
    printf("Case %d: %d
",cnt++,fun(b)-fun(a-1));
}
int main(){
    int loop;memset(dp,-1,sizeof(dp));
   scanf("%d",&loop);
    while(loop--)doit();}