寒假训练数论K

题目
Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.
Input
The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 10 15) and (1 <=N <= 10 9).
Output
For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.
Sample Input
2
1 10 2
3 15 5
Sample Output
Case #1: 5
Case #2: 10
题目大意
给出一个数n,求区间[a,b]内与n互质的数的个数。
解题思路
n的质因子的倍数与n不互质。
对于[1,k]范围内的数,可以先算出质因子倍数的个数,然后用总个数k减去质因子倍数的个数就是[1,k]范围内与n互质的数的个数。
这里用到了容斥原理。
代码

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
using namespace std;
int p[101010],s[101010];
long long f[101010];
bool v[1010100],g[101010];
void work()
{
	memset(f,0,sizeof(f));
	/*
	   容斥原理可以用二进制状态压缩实现
	   例如 1010 表示选了第2个和第四个质数
	*/
	for (int i=1;i<=(1<<s[0])-1;i++)
	  {
	  	 long long num=1,t=0;
	  	 for (int j=0;j<s[0];j++)
	  	   if (i&(1<<j)) 
			 {
			    num*=s[j+1];
			    t++;
			 }
	  	 f[i]=num;  
	  	 if (t%2==0) g[i]=true;
	  	 else g[i]=false;//奇加偶减
	  }
}
void get_s(long long x)//对n进行质因数分解
{
	long long n=x;
	for (int i=2;i*i<=n;i++)
	{
		if (x%i==0)
		{
			s[++s[0]]=i;
			while (x%i==0) x/=i;
		}
	}
	if (x>1) s[++s[0]]=x;
}
long long ask(long long x)
{
	long long ans=0;
	for (int i=1;i<=(1<<s[0])-1;i++)
	  if (g[i]) ans-=(x/f[i]);
	else ans+=(x/f[i]);//奇加偶减
	return x-ans;
}
int main()
{
	freopen("k.in","r",stdin);
	freopen("k.out","w",stdout);
	int T;
	cin>>T;
	for (int k=1;k<=T;k++)
	{
	long long a,b,n;
	cin>>a>>b>>n;
    int t=0;
	s[0]=0;
	get_s(n);
    work();
    long long ans=ask(b)-ask(a-1);
	cout<<"Case #"<<k<<": "<<ans<<endl;  
}   
}