Co-prime + 容斥定理 + 二进制压缩求解

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 816    Accepted Submission(s): 283

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

/*
思路：通常我们求１～ｎ中与ｎ互质的数的个数都是用欧拉函数！ 但如果ｎ比较大或者是求１～ｍ中与ｎ互质的数的个数等等问题，
要想时间效率高的话还是用容斥原理！
容斥、先对n分解质因数，分别记录每个质因数， 那么所求区间内与某个质因数不互质的个数就是n / r(i)，假设r(i)是r的某个质因子
假设只有三个质因子， 总的不互质的个数应该为p1+p2+p3-p1*p2-p1*p3-p2*p3+p1*p2*p3, 及容斥原理，
可以转向百度百科查看相关内容 pi代表n/r(i),即与某个质因子不互质的数的个数 ，当有更多个质因子的时候，
可以用状态压缩解决，二进制位上是1表示这个质因子被取进去了。 如果有奇数个1，就相加，反之则相减
*/
#include<stdio.h>
#include<string.h>
#define MAX 1000
__int64 prime[MAX] ; //用来保存n的质因子
int l = 0 ;
void Cprime(__int64 n) //求n的质因子
{

	for(int i = 2 ; i*i < n ; i ++)
	{
		if( n % i == 0 ){
			prime[l++] = i ;
			while(!(n%i))
			{
				n = n / i ;
			}
		}
	}
	if(n>1)
		prime[l++] = n ;
}
__int64 cal( __int64 x , __int64 n ) //求1-x之间和n互质的数的个数
{
	l = 0 ;
	Cprime(n);
	int i , j ;
	__int64 total = 0 ;
	for( i = 1 ; i < (1<<l) ; 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 
	{
		__int64 cnt = 0 , sum = 1 ;
		for( j = 0 ; j < l ; j ++) //遍历n的所有质因子，判断是否被用到
			if( i & (1<<j))  //这里相当于将i二进制化，然后每位与如果为1，这表达式成立，否则不成立
			{
				sum *=prime[j] ; 
				cnt ++ ;
			}
		if(cnt%2)
			total +=x/sum ;
		else
			total -=x/sum ;
	}
	return x - total ;
}
int main(void)
{
	int t,cnt = 1 ;
	scanf("%d",&t);
	while( t-- )
	{
		__int64 a , b , n ;
		scanf("%I64d %I64d %I64d",&a,&b,&n);
		__int64 sum = cal(b , n ) - cal( a - 1 , n ) ;
		printf("Case #%d: %I64d\n",cnt++ , sum);
	}
	return 0;
}