Investigation LightOJ

An integer is divisible by 3 if the sum of its digits is also divisible by 3. For example, 3702 is divisible by 3 and 12 (3+7+0+2) is also divisible by 3. This property also holds for the integer 9.

In this problem, we will investigate this property for other integers.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains three positive integers A, B and K (1 ≤ A ≤ B < 2³¹ and 0 < K < 10000).

Output

For each case, output the case number and the number of integers in the range [A, B] which are divisible by K and the sum of its digits is also divisible by K.

Sample Input

3

1 20 1

1 20 2

1 1000 4

Sample Output

Case 1: 20

Case 2: 5

Case 3: 64

题解：dp[ pos ][ res1 ][ res2 ]表示当前位置，数位之和模K的余数，这些数位组成的数模K的余数。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int INF=0x3f3f3f3f;

int L,R,K;
int t[10],num[10],dp[10][90][90];

void Inite(){
    t[0]=1;
    for(int i=1;i<10;i++) t[i]=t[i-1]*10;
}

int DFS(int pos,int res1,int res2,bool F){
    if(pos==-1){
        if(res1==0&&res2==0) return 1;
        else return 0;
    }
    
    if(!F&&dp[pos][res1][res2]!=INF) return dp[pos][res1][res2];
    int maxv=F?num[pos]:9;
    
    int ans=0;
    for(int i=0;i<=maxv;i++){
        int x=(res1+i*t[pos])%K;
        int y=(res2+i)%K;
        ans=ans+DFS(pos-1,x,y,(F&&i==maxv)?true:false);
    }
    
    if(!F) dp[pos][res1][res2]=ans;
    return ans;
}

int Solve(int temp){
    if(temp==0) return 1;
    int cnt=0;
    while(temp){
        num[cnt++]=temp%10;
        temp/=10;
    }
    return DFS(cnt-1,0,0,true);
}

int main()
{   Inite();

    int kase;
    cin>>kase;
    for(int k=1;k<=kase;k++){
        cin>>L>>R>>K;
        memset(dp,0x3f,sizeof(dp));
        
        int ans;
        if(K>82) ans=0;
        else ans=Solve(R)-Solve(L-1);
        printf("Case %d: %d
",k,ans);    
    }
    return 0;
}