题目链接：传送门

题意：求区间[a,b]内与n互质的数的个数。

思路：用容斥求出[1-b]与n互质的个数—[1-(a-1)]内与n互质的个数。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cctype>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define maxn 360
#define _ll __int64
#define ll long long
#define INF 0x3f3f3f3f
#define Mod 1000000007
#define pp pair<int,int>
#define ull unsigned long long
#define max(x,y) ( ((x) > (y)) ? (x) : (y) )
#define min(x,y) ( ((x) > (y)) ? (y) : (x) )
using namespace std;
_ll fac[maxn],tot,A,B,N,ans;
void div(int x)
{
	tot=0;
	for(_ll i=2;i*i<=x;i++)
	{
		if(x&&x%i==0)
		{
			fac[tot++]=i;
			while(x&&x%i==0)x/=i;
		}
	}
	if(x>1)fac[tot++]=x;
}
void dfs(_ll num,_ll s,_ll r,_ll n)
{
	if(num==tot)
	{
		if(s&1)ans-=n/r;
		else ans+=n/r;
		return ;
	}
	dfs(num+1,s,r,n);
	dfs(num+1,s+1,r*fac[num],n);
}
int cas=1;
void solve()
{
	div(N);
	ans=0;dfs(0,0,1,B);_ll ans_1_b=ans;
	ans=0;dfs(0,0,1,A-1);_ll ans_1_a=ans;
	printf("Case #%d: %I64d\n",cas++,ans_1_b-ans_1_a);
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%I64d%I64d%I64d",&A,&B,&N);
		solve();
	}
	return 0;
}

HDU 4135-Co-prime（容斥原理）