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): 3549    Accepted Submission(s): 1404

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

21 10 23 15 5

 

Sample Output

Case #1: 5Case #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

 

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#include<stdio.h>  #include<string.h>  #include<stdlib.h>  #include<math.h>  #include<queue>  #include<stack>  #include<algorithm>  #include<iostream>  using namespace std;  long long p[100],k;  void getp(long long n)  {      long long i;      k=0;      for(i=2;i*i<=n;i++)  //保存质因子     {          if(n%i==0)          p[k++]=i;          while(n%i==0)          n/=i;      }      if(n>1) //本身也是质数     p[k++]=n;  }  long long nop(long long m) //求出区间1-m有多少个元素与N不互质 {      long long que[100000],i,j,sum,top=0,t;      que[top++]=-1; //队列数组保存n所有质因子不同组合的乘积     for(i=0;i<k;i++)      {          t=top;          for(j=0;j<t;j++)          {              que[top++]=que[j]*p[i]*(-1); //奇加偶减         }      }      for(i=1,sum=0;i<top;i++)      sum+=m/que[i];  //统计个数     return sum;  }  int main()  {      long long a,b,n;      int t,j=1;      scanf("%d",&t);      while(t--)      {          scanf("%lld%lld%lld",&a,&b,&n);          getp(n);          printf("Case #%d: %lld\n",j++,b-nop(b)-(a-1-nop(a-1)));      }      return 0;  }