Co-prime（容斥定理）

Co-prime

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

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}.

當有n個須要選擇時，一共被選擇的次數減去被選擇兩次的個數，加上被選擇三次的，減去被選擇四次的，依次類推，就是被選擇的並集；
ac代碼：

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

long long t,sum1,sum2,cnt,cnt1,a,b,n,m[105],c=0;

int main(){
	cin>>t;
	while(t--){
		memset(m,0,sizeof(m));
		cin>>a>>b>>n;
		cnt=0;
		//cout<<"%%%"<<cnt<<" "<<n<<endl;
		for(int i=2;i*i<=n;i++){//質因子分解 
			if(n%i==0){//找到n的因子 
				cnt++;
				m[cnt]=i;
			// cout<<" "<<i<<" ";
			}
			//cout<<" N: "<<n<<endl;
			while(n%i==0){
				n/=i;
			} 
			
		}
		//cout<<"ccccnt:"<<cnt<<"n"<<n<<endl;
		if(n!=1){
			cnt++;
			m[cnt]=n;
		// cout<<"NNN:"<<n<<" "<<endl;
		}
		//cout<<"ccnntt:"<<cnt<<endl;
		sum1=0;
		int l=1<<cnt;
		//cout<<"CNT:"<<cnt<<" 1<<cnt : "<<l<<endl;
		for(int i=1;i<(1<<cnt);i++){//1<<cnt即2的cnt次方，即2進制的位數 
			sum2=1;
			cnt1=0;
			for(int j=0;j<cnt;j++){
				int k=1&(i>>j); 
				//cout<<"i:"<<i<<" j:"<<j<<" i>>j:"<<k<<endl; 
				if(1&(i>>j)){//i>>j的意思是將i的二進制向左移j位,按位與1，奇數爲1，偶數爲0 
					sum2*=m[j+1];
					cnt1++;
					//cout<<"i j :"<<i<<" "<<j<<endl; 
				}
			}
			//cout<<"sum2:"<<sum2<<endl;
			if(cnt1&1){
				sum1+=b/sum2;
				sum1-=(a-1)/sum2;
			}
			else{
				sum1-=b/sum2;
				sum1+=(a-1)/sum2;
			}
		}
		cout<<"Case #"<<++c<<": "<<b-a+1-sum1<<endl;
	}
	return 0;
}

/* 2 1 20 12 3 15 5 */