HDU4135(容斥原理)

Posted 2020-08-04 vCoders

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4090    Accepted Submission(s): 1619

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：求1~m中与n互素的数的个数。

#include <cstdio>
#include <vector>
using namespace std;
typedef long long LL;
LL sieve(LL m,LL n)
{
    vector<LL> divisor;
    for(LL i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            divisor.push_back(i);
            while(n%i==0)    n/=i;
        }
    }
    if(n>1)    divisor.push_back(n);
    LL ans=0;
    for(LL mark=1;mark<(1<<divisor.size());mark++)
    {
        LL odd=0;
        LL mul=1;
        for(LL i=0;i<divisor.size();i++)
        {
            if(mark&(1<<i))
            {
                mul*=divisor[i];
                odd++;
            }
        }
        LL cnt=m/mul;
        if(odd&1)    ans+=cnt;
        else ans-=cnt;
    }
    return m-ans;
}
LL a,b,n;
int main()
{
    int T;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        scanf("%lld%lld%lld",&a,&b,&n);
        LL res=sieve(b,n)-sieve(a-1,n);
        printf("Case #%d: ",cas);
        printf("%lld\n",res);
    }
    return 0;
}