hdu4135 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): 626    Accepted Submission(s): 234 

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}.

 

 

Source

The Third Lebanese Collegiate Programming Contest

 

 

Recommend

lcy

 

就是求区间与2互质的数的个数，就是简单容斥吧，用dfs做会比较好。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#define ll long long
using namespace std;

ll ans,l,r,k;
int num,cnt;
ll a[107],prime[100007];
bool boo[100007];
int Case;

void init_prime()
{
    for (int i=2;i<=100000;i++)
    {
        if (!boo[i]) prime[++cnt]=i;
        for (int j=1;j<=cnt&&prime[j]*i<=100000;j++)
        {
            boo[prime[j]*i]=1;
            if (i%prime[j]==0) break;
        }
    }
}
void devide_prime()
{
    num=0;
    for (int i=1;i<=cnt;i++)
    {
        if (k%prime[i]==0)
        {
            a[++num]=prime[i];
            while(k%prime[i]==0) k/=prime[i];
        }
    }
    if (k>1) a[++num]=k;
}
void query(int deep,ll shu,ll qz,ll zhi)
{
    if (deep>num)
    {
        ans+=qz*zhi/shu;
        return;
    }
    query(deep+1,shu*a[deep],-qz,zhi);
    query(deep+1,shu,qz,zhi);
}
int main()
{
    init_prime();
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%lld%lld%lld",&l,&r,&k);
        devide_prime();ans=0;
        query(1,1,1,r);
        query(1,1,-1,l-1);
        printf("Case #%d: %lld\n",++Case,ans);
    }
}