X mod f(x) (思维数位dp)

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

本题不容易想到的是运用枚举进行对所要凑成的数进行枚举

这样就可以在数位dp的过程中进行取余操作

dp数组记录的状态分别是数位 枚举值,当前数位值,余数;

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int dp[11][90][90][90];//数位 枚举值,当前数位值,余数;
int a[11];
int dfs(int pos,int lim, int k,int sum,int mod)
{
    //cout<<pos<<" "<<k<<" "<<sum << " "<<mod<<endl;
    if(pos==0)
    {
        return sum==k&&mod==0;
    }
    if(!lim&&dp[pos][k][sum][mod]!=-1) return dp[pos][k][sum][mod];
    int ans=0,s;
    s=lim?a[pos]:9;
    for(int i=0;i<=s;i++)
    {
        int temp=(mod*10+i)%k;
        ans+=dfs(pos-1,lim&&i==s,k,sum+i,temp);
    }
    if(!lim) dp[pos][k][sum][mod]=ans;
    return ans;
}
int solve(long long x)
{
    int cnt=0;
    while(x!=0)
    {
        ++cnt;
        a[cnt]=x%10;
        x/=10;
    }
    int ans=0;
    for(int i=1;i<=(cnt)*9;i++)
    {
        ans+=dfs(cnt,1,i,0,0);
    }
    return ans;
}
int main()
{
    memset(dp,-1,sizeof(dp));
    int t;
    scanf("%d",&t);
    int cas=0;
    while(t--)
    {
        cas++;
        long long temp1,temp2;
        scanf("%lld%lld",&temp1,&temp2);
        //cout<<solve(temp1)<<" "<<solve(temp2)<<endl;
        printf("Case %d: %d
",cas,solve(temp2)-solve(temp1-1));
    }
}