Co-prime

点击打开链接

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}. 
   
  

本题是求[a,b]中与n互质的数的个数，可以转换成求[1,b]中与n互质的数个数减去[1,a-1]与n互质的数的个数。同时结合二进制；

如 2 3 5 7有 2^4-1 种表示；假设0 0 0 1 说明出现了7；类推此题；

代码：

#include<cstdio>  
#include<algorithm>
using namespace std; 
#define ll long long 
ll p[1000000];  
ll sovle(ll num,int m)  
{  
    ll ans=0,tmp,i,j,flag;  //ans为不是素因子的倍数的数量； 
    for(i=1;i<(ll)(1<<m);i++)  //选数； 
    {  
        tmp=1;//记录选中数字的乘积 
flag=0;  //选中数的数量； 
        for(j=0;j<m;j++)   
            if(i&((ll)(1<<j)))//检测第j个数有没有被选中； 
                {
flag++;
tmp*=p[j];
}
        if(flag&1)//flag%2==1;表示奇数 
            ans+=num/tmp;  
        else  
            ans-=num/tmp;  
    }  
    return ans;  
} 
int main()  
{  
    int T,t=0,m;  
    ll n,a,b,i,k;  
    scanf("%d",&T);  
    while(T--)  
    {  
        scanf("%lld%lld%lld",&a,&b,&n);  
        m=0;  
        for(i=2;i*i<=n;i++) //分解素因子； 
            if(n&&n%i==0)  
            {  
             p[m++]=i;  
             while(n&&n%i==0)  
                    n/=i;  
            }                
        if(n>1)  
            p[m++]=n;  
            k=b-sovle(b,m)-(a-sovle(a-1,m))+1;
        printf("Case #%d: %lld\n",++t,k);  
    }  
    return 0;  
}