C - Co-prime

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.

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这道题分几步：

1、先把n用算术基本原理拆分成不同的质数，如果a~b中的数是这些质数的倍数那其肯定不与n互质，a~b的总个数减去不与n互质的数剩下来就是与n互质的数啦。

2、先计算1~a-1中与n互质的个数，再计算1~b中与n互质的个数，相减。

#include<iostream>
using namespace std;
typedef long long ll;
ll a[];
int main()
{
    ll t;
    cin>>t;
    for(ll ii=;ii<=t;ii++)
    {
        ll x,y,n;
        cin>>x>>y>>n;
        ll num=,sum=,all=,times;
        for(int i=;i*i<=n;i++)   //向a数组里塞n的质因数
        {
             if(n%i==)
            {
                num++;
                a[num]=i;
            }
            while(n%i==)
            {
                n=n/i;
            }
         }
        if(n>)  a[++num]=n;
        for(int i=;i<(<<num);i++)
        {
            times=,sum=;
            for(int j=;j<num;j++)  //num个因数,乘了几个
            {
                if(&(i>>j))      //判断有没有选中第j个因数
                {
                    sum*=a[j+];
                    times++;
                }
            }
            if(sum==||sum==)
            continue;
            if(times%==)
            {
                all+=y/sum;
                all-=(x-)/sum;
            }
            else{
                all-=y/sum;
                all+=(x-)/sum;
            }
        }
        printf("Case #%lld: %lld\n",ii,y-x+-all);
    }
    return ;
} 

以上。