HDU 4135 Co-prime (容斥)

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本文介绍了一种算法，用于计算在给定区间[A, B]内与整数N互质的整数数量。通过寻找N的所有因子并利用容斥原理，文章详细解释了如何高效地解决这一问题。

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Co-prime

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5708 Accepted Submission(s): 2290

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}.

题意

求A到B之间有多少个数和N互质

思路

我们要求互质可以先求A到B之间有几个数和N不互质，和N不互质的数只要是N因子的倍数即可，比如1-10中以2为倍数的个数为 $102=5\frac{10}{2}=5$ ，那么比如N=10，N的因子有2和5，对于1-10范围内，10是2的倍数也是5的倍数，如果我们直接用 $102+105\frac{10}{2}+\frac{10}{5}$ 来计算的话，那么10的数就会被重复计算一次，那我们就要利用容斥原理来剔除重复的情况，容斥的公式如下
[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-QR7oCvtq-1601531427181)(https://gss2.bdstatic.com/9fo3dSag_xI4khGkpoWK1HF6hhy/baike/s=592/sign=3ef1ffbf1fd8bc3ec20806c3b08aa6c8/0ff41bd5ad6eddc403aafe7e3fdbb6fd536633e1.jpg)]
当是为奇数个集合条件时加，偶数个集合条件时减
也就是说比如N=10，1-100中和10不互质的数的个数为 $1002+1005−1002∗5=60\frac{100}{2}+\frac{100}{5}-\frac{100}{2*5}=60$ ,那互质的个数就是100-60=40
所以我们先找出N的因子再利用容斥原理求出与N互质的数的个数再总数减去就行了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int a[10000],num;
void init(int n)
{
    num=0;
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            a[num++]=i;
            while(n%i==0) n/=i;
        }
    }
    if(n>1)
    {
        a[num++]=n;
    }
}
long long ex(long long m)
{
    long long que[10000],sum=0,t=0;
    que[t++]=-1;
    for(int i=0;i<num;i++)
    {
        int k=t;
        for(int j=0;j<k;j++)
            que[t++]=que[j]*a[i]*-1;
    }
    for(int i=1;i<t;i++)
        sum+=m/que[i];
    return sum;
}
int main()
{
    long long a,b;
    int t,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%lld%lld%d",&a,&b,&n);
        printf("Case #%d: ",cas++);
        init(n);
        printf("%lld\n",b-ex(b)-(a-1-ex(a-1)));
    }
    return 0;
}