Light0J-1068-Investigation

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 integerT (≤ 200), Denoting the number of test cases.

Each case contains three positive integersA, BAndK (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 divisibleKAnd the sum of its digits is also divisibleK.

Sample Input	Output for Sample Input
3 1 20 1 1 20 2 1 1000 4	Case 1: 20 Case 2: 5 Case 3: 64

When the array is enabled, dp [12] [10010] [90] will pop up the memory.

Considering that no DP is needed as long as K> = 90 (the sum of single digits cannot exceed 90)

Therefore, you only need to open the array to dp [12] [90] [90].

#include 
 
  #include 
  
   #include 
   
    #include 
    
     using namespace std;typedef long long ll;ll dp[12][90][90];vector
     
       digit;ll m,n,k;ll dfs(int pos,int mod,int sum,int done){    if(pos==-1) return sum==0&&mod==0;    if(!done && ~dp[pos][mod][sum]) return dp[pos][mod][sum];    ll res = 0;    int end = done?digit[pos]:9;    for(int i = 0; i <= end; i++){        res += dfs(pos-1,(mod*10+i)%k,(sum+i)%k,done&&i==end);    }    if(!done) dp[pos][mod][sum] = res;    return res;}ll solve(ll x){    digit.clear();    while(x){        digit.push_back(x%10);        x /= 10;    }    return dfs(digit.size()-1,0,0,1);}int main(){    int ncase,T=1;    cin >> ncase;    while(ncase--){       cin >> m >> n >> k;       if(k >= 90){            printf("Case %d: 0\n",T++);            continue;       }       memset(dp,-1,sizeof dp);       printf("Case %d: %d\n",T++,solve(n)-solve(m-1));    }    return 0;}