HDU 4135 Co-prime(容斥原理)

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3313    Accepted Submission(s): 1286

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

求在n在a~b之间的和n互质的个数

容斥原理模板题
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>

using namespace std;
typedef long long int LL;
const LL INF=(LL)1<<62;
#define MAX 1000000
LL prime[MAX+5];
LL sprime[MAX+5];
LL q[MAX+5];
bool check[MAX+5];
LL n,a,b,cnt;
void eular()
{
    memset(check,false,sizeof(check));
    int tot=0;
    for(LL i=2;i<MAX+5;i++)
    {
        if(!check[i])
            prime[tot++]=i;
        for(int j=0;j<tot;j++)
        {
            if(i*prime[j]>MAX+5) break;
             check[i*prime[j]]=true;
            if(i%prime[j]==0) break;
        }
    }
}
void Divide(LL n)
{
    cnt=0;
    LL t=(LL)sqrt(1.0*n);
    for(LL i=0;prime[i]<=t;i++)
    {
        if(n%prime[i]==0)
        {
            sprime[cnt++]=prime[i];
            while(n%prime[i]==0)
                n/=prime[i];
        }
    }
    if(n>1)
        sprime[cnt++]=n;
}
LL Ex(LL n)
{
    LL sum=0,t=1;
    q[0]=-1;
    for(LL i=0;i<cnt;i++)
    {
        LL x=t;
        for(LL j=0;j<x;j++)
        {
            q[t]=q[j]*sprime[i]*(-1);
            t++;
        }
    }
    for(LL i=1;i<t;i++)
        sum+=n/q[i];
    return sum;
}
int main()
{
    int t;
    scanf("%d",&t);
    eular();
    int cas=0;
    while(t--)
    {
        scanf("%lld%lld%lld",&a,&b,&n);
        Divide(n);
        printf("Case #%d: %lld
",++cas,(b-Ex(b)-(a-1-Ex(a-1))));
    }
    return 0;
}