HDU 4389 数位dp

X mod f(x)

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2992    Accepted Submission(s): 1171

Problem 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 <= 10⁹), 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

 

Author

WHU

 

Source

2012 Multi-University Training Contest 9

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
#include <map>
#define ll  __int64
#define dazhi 2147483647
#define bug() printf("!!!!!!!")
#define M 100005
using namespace  std;
int bit[10];
int dp[10][82][82][82];
int  n;
int  functi(int pos,int mod,int xx ,int sum,bool flag)
{
    if(pos==0) return (xx==sum&&mod%sum==0);
    if(flag&&dp[pos][mod][xx][sum]!=-1) return dp[pos][mod][xx][sum];
    ll x=flag ? 9 : bit[pos];
    ll ans=0;
    for(ll i=0;i<=x;i++)
    ans+=functi(pos-1,(mod*10+i)%xx,xx,sum+i,flag||i<x);
    if(flag)
        dp[pos][mod][xx][sum]=ans;
    return ans;
}
int  fun(int x)
{
    int len=0;
    while(x)
    {
        bit[++len]=x%10;
        x/=10;
    }
    ll re=0;
    for(int i=1;i<=81;i++)
        re+=functi(len,0,i,0,0);
    return re;
}
int main()
{
   int t;
   int  l,r;
   while(scanf("%d",&t)!=EOF)
   {
       memset(dp,-1,sizeof(dp));
       for(int i=1;i<=t;i++)
       {
           scanf("%d %d",&l,&r);
           int exm=fun(r);
           printf("Case %d: %d
",i,exm-fun(l-1));
       }
   }
    return 0;
}