hdu4135 Co-prime(分解质因数，容斥)

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4433    Accepted Submission(s): 1755

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 <= 10 ¹⁵) and (1 <=N <= 10 ⁹).

 

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

 

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题意：[A，B]区间内，与N互质的数的个数。可以把问题转换为[1，B]区间的个数减去[1,A-1]区间的个数。现在的问题就是如何求[1,m]中与n互质的数的个数。可求出[1,m]区间与n不互质的数的个数，假设为num，那么我们的答案就是：m-num。然后问题就是如何求[1,m]区间与n不互质的数的个数。

可以举例说明

m=12,n=30.

第一步：求出n的质因子：2,3,5；

第二步：(1,m)中是n的因子的倍数当然就不互质了(2,4,6,8,10,12)->n/2  6个      (3,6,9,12)->n/3  4个,      (5,10)->n/5  2个。

答案显然不是全加起来。里面明显出现了重复的,我们现在要处理的就是如何去掉那些重复的了！

第三步：这里就需要用到容斥原理了，公式就是：n/2+n/3+n/5-n/(2*3)-n/(2*5)-n/(3*5)+n/(2*3*5)。发现除数是奇数个的时候是加，偶数个的时候是减。

#include<iostream>
using namespace std;
long long a[1000],num;
long long mod=1e9+7;
void init(long long n)//分解质因数 
{
    long long i;
    num=0;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            a[num++]=i;
            while(n%i==0)
                n=n/i;
        }
    }
    if(n>1)
        a[num++]=n;
}
long long rongchi(long long m)
{
    long long que[10000],i,j,k,sum=0;
    int t=0;
    que[t++]=-1;
    for(i=0;i<num;i++)
    {
        k=t;
        for(j=0;j<k;j++)
           que[t++]=que[j]*a[i]*(-1);
    }
    for(i=1;i<t;i++)
        sum=sum+m/que[i];
    return sum;
}
int main()
{
    int t;
   	int cas=0;
    long long a,b,n;
    scanf("%d",&t);
    while(t--)
    {
        //cout<<le<<" "<<ri<<endl;
        long long aans=0;
        //scanf("%d%d%d",&a,&b,&n);
        cin>>a>>b>>n;
        init(n);
       // cout<<a<<endl;
        aans=b-rongchi(b)-(a-1-rongchi(a-1));
        //cout<<1<<endl;
        printf("Case #%d: ",++cas);
        printf("%lld\n",aans);
    } 
}