# Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4289    Accepted Submission(s): 1698

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

AC代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
typedef __int64 LL;
using namespace std;
vector<LL>prime;
void jprime(LL n)   //分解质因子
{
for(LL i=2; i*i<=n; i++)
if(n%i==0)
{
prime.push_back(i); //质因子
while(n%i==0)
n/=i;
}
if(n>1)prime.push_back(n);
}

LL solve(LL a,LL b,LL n)    //[a,b]中与n互质的数的个数
{
LL sum=0;
LL size=prime.size();
for(LL msk=1; msk<(1<<size); msk++) //共有2^size种组合情况
{
LL mult=1,bits=0;
for(LL i=0; i<size; i++)    //枚举每一种情况
{
if(msk&(1<<i))  //如果在当前判断之内
{
++bits;
mult*=prime[i];
}
}
LL cur=b/mult-(a-1)/mult;   //中间结果
if(bits&1)sum+=cur;
else sum-=cur;
}
return b-a-sum+1;
}
int main()
{
LL T;
scanf("%I64d",&T);
for(LL t=1; t<=T; t++)
{
LL a,b,n;
scanf("%I64d%I64d%I64d",&a,&b,&n);
prime.clear();
jprime(n);
printf("Case #%I64d: %I64d\n",t,solve(a,b,n));
}
return 0;
}