uva 11361

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
The rst line of input is an integer
T(T<100) that indicates the number of test cases. Each case is a line containing 3 positive integers A,B and K.
1<=A<=B<=2^31 and 0<K<10000.
Output
For each case, output the number of integers in the range [
A;B] which is divisible by K and the sum of its digits is also divisible by K.
SampleInput
3
1 20 1
1 20 2
1 1000 4
SampleOutput
20
5
64

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <algorithm>

using namespace std;

const int ten[] = {1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};

int f[11][90][90];
vector<int> v[2];
int a,b,k;

void dev(int n,int index){
if (n == 0){
v[index].push_back(0);
return ;
}

while (n){
v[index].push_back(n % 10);
n /= 10;
}
}

int fun(const int &index,int sum,int m,int pos){

if (pos < 0) return 0;

int ans = 0;
if (pos == 0){
for (int i = 0; i <= v[index][pos]; ++i){
if ((sum + i) % k == 0 && (m * 10 + i) % k == 0){
++ans;
}
}
return ans;
}

for (int i = 0; i < v[index][pos]; ++i){
ans += f[pos][(k - (sum + i) % k) % k][(k - (m * 10 + i) * ten[pos] % k) % k];
}

return ans + fun(index,sum + v[index][pos],(m * 10 + v[index][pos]),pos-1);
}

int main(){

int T;
scanf("%d",&T);

while (T--){
scanf("%d%d%d",&a,&b,&k);
if (k > 81){
printf("0\n");
continue;
}
memset(f,0,sizeof(f));

for (int i = 0; i < 10; ++i){
f[1][i % k][i % k]++;
}

for (int i = 1; i <= 9; ++i){
for (int m1 = 0; m1 < k; ++m1){
for (int m2 = 0; m2 < k; ++m2){

for (int x = 0; x < 10; ++x){
f[i][m1][m2] += f[i-1][(m1-x%k+k)%k][(m2-ten[i-1]*x%k+k)%k];
}
}
}
}

for (int i = 0; i < 2; ++i){
v[i].clear();
}

dev(b,0);
dev(a-1,1);

printf("%d %d\n",fun(0,0,0,v[0].size()-1) , fun(1,0,0,v[1].size()-1));
}

return 0;
}