容斥原理详解 以及代码的剖析 结合实例hdu4135

原理：首先考虑一个问题，1000以内6,7,8,9的倍数有多少个？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)

这是容斥原理的一个最简单的应用，类比这道题，Step3到4其实将每个数a的不重复约数记录了下来，有公共约数的四个数的方案要从ans中减去，多减的要加上

奇加偶减

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 <= 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}. 
    
   
   
   
   

 

题意： 求 a到b的数中与n互质的数的个数

代码是参考人家的

参考地址

http://blog.csdn.net/duanxian0621/article/details/7303767

#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中的所有数与m个质因子不互质的个数 注意是不互质哦
{
	LL ans=0,tmp,i,j,flag;
	for(i=1;i<(LL)(1<<m);i++)
	{ //用二进制来1,0来表示第几个素因子是否被用到,如m=3，三个因子是2,3,5，则i=3时二进制是011，表示第2、3个因子被用到 
		tmp=1,flag=0;
		for(j=0;j<m;j++) 
			if(i&((LL)(1<<j)))//判断第几个因子目前被用到
				flag++,tmp*=prime[j];
		if(flag&1)//容斥原理，奇加偶减 
			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++) //对n进行素因子分解 
			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;
}