HDU 4135：Co-prime（容斥+二进制拆分）

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6869    Accepted Submission(s): 2710
 

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

给出三个数a,b,n，求区间[a.b]中有多少和n互质的数

思路

先把n的质因子记录下来，然后利用容斥+二进制拆分分解求出1~（a-1）和1~b之间的与n互质的个数ans1和ans2，然后减去区间中数的个数减去（ans2-ans1）即可

AC代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const double E=exp(1);
const int maxn=1e6+10;
using namespace std;
int A[maxn];//用来存放质因子
ll gcd(ll a,ll b)
{
	return b?gcd(b,a%b):a;
}
ll lcm(ll a,ll b)
{
	return a/gcd(a,b)*b;
}
int main(int argc, char const *argv[])
{
	int t;
	ll a,b,n;
	scanf("%d",&t);
	int x=0;
	while(t--)
	{
		ll ans1,ans;
		ans=ans1=0;
		map<int,int>mmp;//记录质因子是否出现过
		scanf("%lld%lld%lld",&a,&b,&n);
		ll m=n;
		int k=0;
		for(int i=2;i*i<=m;i++)
		{
			if(m%i==0)
			{
				while(m%i==0)
				{
					if(mmp[i]==0)
					{
						A[k++]=i;
						mmp[i]=1;
					}
					m/=i;
				}
			}
		}
		if(m>1)
		{
			A[k++]=m;
			mmp[m]=1;
		}
		for(int i=1;i<(1<<k);i++)
		{
			int cnt=0;
			ll tmp=1;
			for(int j=0;j<k;j++)
			{
				if(i>>j&1)
				{
					cnt++;
					tmp=lcm(tmp,A[j]);
				}
			}
			if(cnt&1)
			{
				ans+=(a-1)/tmp;
				ans1+=(b)/tmp;
			}
			else
			{
				ans-=(a-1)/tmp;
				ans1-=b/tmp;
			}
		}
		printf("Case #%d: %lld
",++x,(b-a+1)-ans1+ans);
	}
	return 0;
}