容斥定理

最新推荐文章于 2021-07-31 15:35:51 发布

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最新推荐文章于 2021-07-31 15:35:51 发布

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文章标签：容斥定理

本文链接：https://blog.youkuaiyun.com/qq_44797733/article/details/99655089

本文介绍了一种算法，用于计算在给定范围内与特定数N互为共轭的整数数量。通过分解N的质因数，利用容斥原理计算[A,B]区间内与N互质的数。输入包括测试用例数量及每个案例的范围和目标数N。

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}.

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
int a[20];
int main()
{
	int t;
	cin>>t;
	for(int i=1;i<=t;i++){
		ll A,B,n;
		ll sum=0;
		scanf("%lld %lld %lld",&A,&B,&n);
		int cnt=0;
		for(int i=2;i*i<=n;i++){
			if(n%i==0) a[++cnt]=i;
			while(n%i==0){
				n/=i;
			}
		}
		if(n>1){
			a[++cnt]=n;
		}
		for(int i=1;i<(1<<cnt);i++){
			int cnt1=0;
			ll sum1=1;
			for(int j=0;j<cnt;j++){
				if(1&(i>>j)){
					cnt1++;
					sum1*=a[j+1];//求一个集合的并集 有公约数2,3,5  3个集合的并集 
				}
			}
			if(cnt1&1){
				sum+=B/sum1;
				sum-=(A-1)/sum1;
			}
			else{
				sum-=B/sum1;
				sum+=(A-1)/sum1;
			}
		}
		printf("Case #%d: %lld\n",i,(B-A+1)-sum);
	}
	return 0;
}

找一个数的质因数即可一个大于1的自然数可分为质数的积找到A和B之间能整除n的质因子的部分然后容斥定理即可