原理：首先考慮一個問題，1000以內6,7,8,9的倍數有多少個？答案是

1000div6+1000div7+1000div8+1000div9

-1000div(6*7)-1000div(6*8)-1000div(6*9)-1000div(7*8)-1000div(7*9)-1000div(8*9)

+1000div(6*7*8)+1000div(6*8*9)+1000div(7*8*9)

-1000div(6*7*8*9)

這是容斥原理的一個最簡單的應用，類比這道題，Step3到4其實將每個數a的不重複約數記錄了下來，有公共約數的四個數的方案要從ans中減去，多減的要加上

奇加偶減

hdu 4135

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 87    Accepted Submission(s): 39

Problem Description 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 <= 10¹⁵) and (1 <=N <= 10⁹). 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 HintIn the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.

題意： 求 a到b的數中與n互質的數的個數

程式碼是參考人家的

參考地址

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define LL long long 
#define maxn 70

LL prime[maxn];
LL make_ans(LL num,int m)//1到num中的所有數與m個質因子不互質的個數 注意是不互質哦
{
	LL ans=0,tmp,i,j,flag;
	for(i=1;i<(LL)(1<<m);i++)
	{ //用二進位制來1,0來表示第幾個素因子是否被用到,如m=3，三個因子是2,3,5，則i=3時二進位制是011，表示第2、3個因子被用到 
		tmp=1,flag=0;
		for(j=0;j<m;j++) 
			if(i&((LL)(1<<j)))//判斷第幾個因子目前被用到
				flag++,tmp*=prime[j];
		if(flag&1)//容斥原理，奇加偶減 
			ans+=num/tmp;
		else
			ans-=num/tmp;
	}
	return ans;
}

int main()
{
	int T,t=0,m;
	LL n,a,b,i;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%I64d%I64d%I64d",&a,&b,&n);
		m=0;
		for(i=2;i*i<=n;i++) //對n進行素因子分解 
			if(n&&n%i==0)
			{
				prime[m++]=i;
				while(n&&n%i==0)
					n/=i;
			}	
		if(n>1)
			prime[m++]=n;
		printf("Case #%d: %I64d\n",++t,(b-make_ans(b,m))-(a-1-make_ans(a-1,m)));
	}
	return 0;
}