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)
Total Submission(s): 3619    Accepted Submission(s): 1409

 

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

 

#include<cstdio>
#include<cstring>
int s[11];int dp[11][82][82][82];

int dfs(int pos,int sum,int divisor,int reminder,bool lim)
{
    if(pos<=0)return sum==divisor&&reminder==0;
    if(!lim&&dp[pos][sum][divisor][reminder]!=-1)return dp[pos][sum][divisor][reminder];
    int num=lim?s[pos]:9;int ans=0;
    for(int i=0;i<=num;i++)
    {
        int sum_x=sum+i;
        ans+=dfs(pos-1,sum_x,divisor,(reminder*10+i)%divisor,lim&&i==num);
    }
    return lim?ans:dp[pos][sum][divisor][reminder]=ans;
}

int solve(int x)
{
    int ans=0;
    int t=0;
    while(x){
        s[++t]=x%10;
        x/=10;
    }
    for(int i=1;i<=81;i++)
        ans+=dfs(t,0,i,0,1);
    return ans;
}

int main()
{
    int t,c=1;
    int n,m;scanf("%d",&t);memset(dp,-1,sizeof(dp));
    while(t--){
        scanf("%d%d",&n,&m);
        printf("Case %d: %d\n",c++,solve(m)-solve(n-1));
    }
}