先把n素因子分解，再用位运算，枚举倍数，加上容斥原理

#include<stdio.h>
#include<vector>
using namespace std;
#define ll long long
ll work(ll r, ll n){
	vector<ll> p;
	for(ll i = 2; i * i <= n; i++){
		if(n % i==0){
			p.push_back(i);
			while(n%i==0) n /= i;
		}
	}
	if(n>1) p.push_back(n);
	ll ans = 0;
	for(ll m = 1; m < (1<<p.size()); m++){
		ll mul = 1, bits = 0;
		for(ll i = 0; i < p.size(); i++){
			if(m & (1<<i)){	
				++bits;
				mul *= p[i];
			}
		}
		ll cur = r / mul;
		if(bits & 1) ans += cur;
		else ans -= cur;
	}
	return r - ans;
}
int main()
{
	//freopen("in.txt","r",stdin);
	int t;
	ll n, l, r;
	scanf("%d", &t);
	int ca = 1;
	while(t--)
	{
		scanf("%lld%lld%lld", &l, &r, &n);
		printf("Case #%d: %lld\n", ca++, work(r, n) - work(l-1, n));
	}
	return 0;
}