HDU 4135 Co-prime（容斥+数论）

Posted 2020-10-03 勿忘初心0924

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5526    Accepted Submission(s): 2209

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

 

/*
* @Author: lyuc
* @Date:   2017-08-16 16:52:21
* @Last Modified by:   lyuc
* @Last Modified time: 2017-08-16 21:44:34
*/

/*
 题意：给你a,b,n让你求在区间[a,b]内有多少数与n互质

 思路：求出n的所有质因子，然后[1,a]区间内的与n互质的数的数量就是，a减去不互质的数的数量，同理
     [1,b]的也可以这么求,容斥求出这部分的结果
*/
#include <bits/stdc++.h>

#define LL long long
#define MAXN 1005
#define MAXM 10005

using namespace std;

int t;
LL a,b,n;
LL factor[MAXN];
LL tol;
LL que[MAXM];

void div(LL n){//筛出来n的因子
    tol=0;
    for(LL i=2;i*i<=n;i++){
        if(n%i==0){
            factor[tol++]=i;
            while(n%i==0){
                n/=i;
            }
        }
    }
    if(n!=1) factor[tol++]=n;//如果是素数的话加上本身
}

LL cal(LL x){//容斥计算出[1,x]内与n不互质的元素的个数
    LL res=0;
    LL t=0;
    que[t++]=-1;
    for(int i=0;i<tol;i++){
        int k=t;
        for(int j=0;j<k;j++){
            que[t++]=que[j]*factor[i]*-1;
        }
    }
    for(int i=1;i<t;i++){
        res+=x/que[i];
    }
    return res;
}

int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++){
        printf("Case #%d: ",ca);
        scanf("%lld%lld%lld",&a,&b,&n);
        div(n);
        printf("%lld\n",b-cal(b)-(a-1-cal(a-1)) );
    }
    return 0;
}