hdu 4135 Co-prime(容斥原理)

Co-prime

 

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

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}.

不懂容斥原理的话，这有容斥原理的上手题： 点击打开链接

有两个原理：

1. 在[1,n]之间与m互质的数的个数=[1,n]之间的总个数-[1,n]之间与m不互质的数的个数

2. [1,n]之间与m互质的数的数量 = n - （包含一个质因子的数的个数）+（包含2个质因子的数的个数）-（包含3个质因子的数的个数）+（包含4个质因子的数的个数）.....

代码：

#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;

#define LL long long int

vector<int>q;
LL b[1000];

void div(int  k)//对k分解质因子
{
    for(int i=2;i*i<=k;i++)
    {
        if(k%i==0)
            q.push_back(i);
        while(k%i==0)
            k/=i;
    }
    if(k>1)
        q.push_back(k);
}
//容斥原理计算[1,n]内有多少个与k互质的数
LL cont(LL n)
{
    LL g=0,sum=n;
    b[++g]=1;
    for(LL i=0;i<q.size();i++)
    {
        LL t=g;
        for(LL j=1;j<=g;j++)
        {
            b[++t]=-q[i]*b[j];
            sum+=n/b[t];
        }
        g=t;
    }
    return sum;
}

int main()
{
    int t,m=0;
    scanf("%d",&t);
    while(m<t)
    {
        q.clear();
        LL a,b;
        int k;
        scanf("%lld%lld%d",&a,&b,&k);
        div(k);
        printf("Case #%d: %lld\n",cont(b)-cont(a-1));
    }
    return 0;
}