hdu 4389 X mod f(x) 数位dp

X mod f(x)

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

dp进阶之路写法；

或者分块打表；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
int f[11][83][83][83],bit[11];
int dp(int pos,int sum,int m,int p,int flag)
{
    if(pos==0)return (sum==p&&m==0);
    if(flag&&f[pos][sum][m][p]!=-1)return f[pos][sum][m][p];
    int x=flag?9:bit[pos];
    int ans=0;
    for(int i=0;i<=x;i++)
    {
        ans+=dp(pos-1,sum+i,(m*10+i)%p,p,flag||i<x);
    }
    if(flag)f[pos][sum][m][p]=ans;
    return ans;
}
int getans(int x,int p)
{
    int len=0;
    while(x)
    {
        bit[++len]=x%10;
        x/=10;
    }
    return dp(len,0,0,p,0);
}
int main()
{
    int T,cas=1;
    memset(f,-1,sizeof(f));
    scanf("%d",&T);
    while(T--)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        int ans=0;
        for(int i=1;i<=81;i++)
            ans+=getans(r,i)-getans(l-1,i);
        printf("Case %d: %d
",cas++,ans);
    }
    return 0;
}