HDU 4135(基本的容斥定理)

              

                                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.

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

题意：给定a、b、c，求a到b区间内与c互质的数。

思路：

通常我们求１～ｎ中与ｎ互质的数的个数都是用欧拉函数！ 但如果ｎ比较大或者是求１～ｍ中与ｎ互质的数的个数等等问题， 要想时间效率高的话还是用容斥原理！

容斥、先对n分解质因数，分别记录每个质因数， 那么所求区间内与某个质因数不互质的个数就是n / r(i)，假设r(i)是r的某个质因子 假设只有三个质因子， 总的不互质的个数应该为p1+p2+p3-p1*p2-p1*p3-p2*p3+p1*p2*p3, 及容斥原理，可以转向百度百科查看相关内容 pi代表n/r(i),即与某个质因子不互质的数的个数 ，当有更多个质因子的时候， 可以用状态压缩解决，二进制位上是1表示这个质因子被取进去了。 如果有奇数个1，就相加，反之则相减.

#include
#include
#include
using namespace std;
vectorx;
void divide(int n)///素因子的分解
{
    x.clear();
    for(int i=2;i*i<=n;i++)
        if(n%i==0)
    {
        x.push_back(i);
        while(n%i==0)
        {
            n/=i;
        }
    }
    if(n!=1)
        x.push_back(n);
}
long long solve(long long b,long long n)
{
    long long sum=0;
    for(int i=1;i<(1<>t;
    long long a,b,n;
    for(int i=1;i<=t;i++)
    {
        cin>>a>>b>>n;
        if(n==0)
        {
            cout<<"Case #"<