hdu 4135 a到b的范围中多少数与n互质（容斥）

Co-prime

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4135

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).

题解：先求出n的所有质因数，因为n最大为1e9所以最多10个

利用二进制来模拟是否乘上某个质因数，例如有个n为2*3*5=30

1-x中共有ans个数与其互质  ans=-x/2-x/3-x/5+x/6+x/10+x/15-x/30

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
long long fa[20];
long long factor(long long x)
{
    long long num=0;
    for(long long i=2;i*i<=x;i++)
    {
         if(x%i==0)
         {
             fa[num++]=i;
             while(x%i==0)x/=i;
         }
    }
    if(x>1)fa[num++]=x;
    return num;
}
long long un(long long  x,long long num)
{
    long long res=0;
    for(long long i=0;i<(1<<num);i++)
    {
        long long g=1,k=0;
        for(long long j=0;j<num;j++)
        {
            if(i&(1<<j))
              g*=fa[j],k++;
        }
        if(k%2)
            res-=x/g;
        else
            res+=x/g;
    }
    return res;
}
int main()
{
    long long T,n,i=1;
    long long a,b,ans;
    scanf("%I64d",&T);
    while(T--)
    {
        scanf("%I64d %I64d %I64d",&a,&b,&n);
        long long num=factor(n);
        ans=un(b,num)-un(a-1,num);
        printf("Case #%I64d: ",i++);
        printf("%I64d\n",ans);
    }
	return 0;
}