hdu4135 容斥定理

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3587    Accepted Submission(s): 1419

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}.

#include<stdio.h>
#include<string.h>
using namespace std;
long long a[],num;
void prime(long long n)
{
    long long i;
    num=;
    for(i=;i*i<=n;i++)
    {
        if(n%i==)
        {
            a[num++]=i;
            while(n%i==)
                n=n/i;
        }
    }
    if(n>)
        a[num++]=n;
}
long long get(long long m)
{
    long long que[],k,t=,sum=;
    que[t++]=-;
    for(int i=;i<num;i++)
    {
        k=t;
        for(int j=;j<k;j++)
           que[t++]=que[j]*a[i]*(-);
    }
    for(int i=;i<t;i++)
        sum=sum+m/que[i];
    return sum;
}
 
int main()
{
    long long T,x,y,n,cnt;
    while(scanf("%lld",&T)!=EOF)
    {
        for(int i=;i<=T;i++)
        {
           scanf("%lld%lld%lld",&x,&y,&n);
           prime(n);
           cnt=y-get(y)-(x--get(x-));
           printf("Case #%d: ",i);
           printf("%lld\n",cnt);
        }
    }
    return ;
}