2
1 10 2
3 15 5

Case #1: 5
Case #2: 10

Hint
In the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}. 
 

#include <cstdio>
#include <cstring>
#define ll long long
int const MAX = 5e5 + 5;
int fac[MAX];
int cnt;
ll a, b, n, num1, num2;

void get_factor(ll x)
{
    cnt = 0;
    for(int i = 2; i * i < MAX; i++)
    {
        if(x % i == 0)
        {
            fac[cnt ++] = i;
            while(x % i == 0)
                x /= i;
        }
    }
    if(x > 1)
        fac[cnt ++] = x;
}   

void DFS(int idx, ll cur, int sgin, ll s, bool f)
{
    for(int i = idx; i < cnt; i++)
    {
        ll tmp = (ll) cur * fac[i];
        if(f)
            num1 += (ll) sgin * (s / tmp);
        else
            num2 += (ll) sgin * (s / tmp);
        DFS(i + 1, tmp, -sgin, s, f);
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    for(int ca = 1; ca <= T; ca++)
    {
        scanf("%I64d %I64d %I64d", &a, &b, &n);
        get_factor(n);
        num1 = 0;
        DFS(0, 1, 1, b, 1);
        num2 = 0;
        DFS(0, 1, 1, a - 1, 0);
        ll ans = (ll)b - a + 1 - (num1 - num2);
        printf("Case #%d: %I64d\n", ca, ans);
    }
}