HDU 4135 Co-prime (包含排斥原理+筛选法求素数)

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3216    Accepted Submission(s): 1245

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

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：给定一个区间[a,b]，求该区间内与n互素的数，（互素：一个数x与n没有相同的因子，除1外）。

思路：由于区间范围太大，遍历每个数肯定不行。求区间[a,b]内的所有与n互素的数，即可以转化为求区间[1,b]的所有与n互素的数减去[1,a-1]的所有与n互素的数。先求出n的所有质因子，存入数组subprime[i]。b/subprime[i]  即为[1,b]中能被subprime[i]整除的数的数量。这里就用到包含排斥原理。

两个集合时： |A∪B| =|A∪B| = |A|+|B| - |A∩B |

三个集合时：|A∪B∪C| = |A|+|B|+|C| - |A∩B| - |B∩C| - |C∩A| + |A∩B∩C|

多个时，以此类推，加减循环。

#include
#include
#include
#include
#include
#include
#define MAXN 100005
using namespace std;
bitsetisprime;
long long prime[MAXN];
long long subprime[MAXN];
long long a[MAXN];
long long cnt,cnt2;
void is_prime(){//筛选法求素数，时间复杂度为O(n) 
	isprime.set();//把isprime中的所有数都置为1 
	long long i,j;
	cnt=0;
	isprime[1]=0;
	for(i=2;i<=MAXN;i++){
		if(isprime[i]){
			prime[cnt++]=i;
			for(j=i+i;j<=MAXN;j+=i){
				isprime[j]=0;
			}
		}
	}
}
void f(long long n){//分解n的质因数 
	long long i,j;
	cnt2=0;
	for(i=0;prime[i]<=sqrt(n*1.0);i++){
		if(n%prime[i]==0){
			subprime[cnt2++]=prime[i];
			while(n%prime[i]==0){//反复除去相同质因子，到最后一定会只剩下1或者一个大于sqrt(n)的素数 
				n/=prime[i];
			}
		}
	}
	if(n>=2){//如果留下了一个大于2的数，那这个数一定是素数，而且是它的因子 
		subprime[cnt2++]=n;
	}
}
long long solve(long long x){//包含排斥原理 
	long long i,j,cur,k,sum=0;
	a[0]=-1;
	k=1;
	for(i=0;i>t;
	is_prime();
    cases=1;
	while(t--){
		cin>>a>>b>>n;
		f(n);
		ans=(b-solve(b))-(a-1-solve(a-1));
		printf("Case #%d: ",cases++);
		cout<