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.

数组开dp[12][10010][90] 就会爆内存

考虑到只要K>=90 就不用DP了(个位数之和不会超过90)

因此只要将数组开到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; }