HDU 4135 Coprime (容斥原理)

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 626    Accepted Submission(s): 234

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}.

 

Source

The Third Lebanese Collegiate Programming Contest

 

Recommend

lcy

 

 

 

 

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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>

using namespace std;

long long a,b,n;
vector<long long> vt;

long long solve(long long x,long long n){
    vt.clear();
    long long i,j;
    for(i=2;i*i<=n;i++)     //对n进行素数分解
        if(n%i==0){
            vt.push_back(i);
            while(n%i==0)
                n/=i;
        }
    if(n>1)
        vt.push_back(n);

    long long sum=0,val,cnt;
    for(i=1;i<(1<<vt.size());i++){  //用二进制来1,0来表示第几个素因子是否被用到,如m=3，三个因子是2,3,5，则i=3时二进制是011，表示第2、3个因子被用到
        val=1;
        cnt=0;
        for(j=0;j<vt.size();j++)
            if(i&(1<<j)){       //判断第几个因子目前被用到 
                val*=vt[j];
                cnt++;
            }
        if(cnt&1)       //容斥原理，奇加偶减
            sum+=x/val;
        else
            sum-=x/val;
    }
    return x-sum;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t,cases=0;
    scanf("%d",&t);
    while(t--){
        cin>>a>>b>>n;
        cout<<"Case #"<<++cases<<": "<<solve(b,n)-solve(a-1,n)<<endl;
    }
    return 0;
}