Co-prime HDU - 4135

最新推荐文章于 2020-06-26 18:16:11 发布

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最新推荐文章于 2020-06-26 18:16:11 发布

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分类专栏：容斥文章标签：容斥

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本文介绍了一种解决HDU_4135Co_prime问题的方法，该问题要求计算在指定区间内与给定整数互质的所有整数的数量。通过分解给定整数的素因子，并利用容斥原理来计算结果。

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HDU _4135 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.

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).

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.

Input

2
1 10 2
3 15 5

Output

Case #1: 5
Case #2: 10

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：给定区间<a,b> 求在区间a,b里与 n 互质的数有多少个。

思路：大概算一个模板吧：首先考虑欧拉函数，明显不行，数据太大。那么就考虑先算出1~a 中与 n 互质的数，再算1~b 中与 n 互质的数，然后ans= solve(b)-solve(a);

问题就变为了 1~a 中与 n 互质的数怎么求，博客上COPY 的思路：

对于 1~ a ,我们先考虑与 n 不互质的数，然后互质的数就是 a-Solve(a);

对于不互质的数：先分解 n 找到素因子，不互质的数就是 a 以内素因子的倍数的个数。

举个例子：

n= 20 素因子 ={ 2，5 }

a=15 那么不互质的个数就是 15/2+15/5 ={2，4，6，8，10，12，14}+{5，10，15}=10-1=9；

那么对于素因子存在上例中 {10} 这个数在两个集合中都出现，此时就需要用到容斥原理。

然后就是容斥的实现了。

容斥的实现有些技巧，便于理解我给出一个每步的进行，

以 1 12 30 为例子。

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include <bits/stdc++.h>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#define maxn 1001000
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MOD 1000000007
#define ll long long
using namespace std;


ll Q[maxn],fac[maxn];
ll num;
void Divide(ll n)
{
    num=0;
    for(ll i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            while(n%i==0)
                n/=i;
            fac[num++]=i;
        }
    }
    if(n!=1) fac[num++]=n;
}
ll solve(ll n)
{
    ll k,t,ans;
    t=ans=0;
    Q[t++]=-1;
    for(ll i=0;i<num;i++)
    {
       // cout<<"num "<<num<<endl;
        k=t;
        for(ll j=0;j<k;j++)
        {
            Q[t++]=-1*Q[j]*fac[i];
           // printf("fac[%lld]->%lld  -- Q[%lld] -> %lld\n",i,fac[i],t-1,Q[t-1]);
        }
    }
    for(ll i=1;i<t;i++)
    {
        ans+=n/Q[i];
    }
    return ans;
}

int main()
{
    int T,kase=1;
    scanf("%d",&T);
    while(T--)
    {
        ll a,b,n;
        scanf("%lld%lld%lld",&a,&b,&n);
        Divide(n);
        ll ans=b-solve(b)-(a-1-solve(a-1));
        printf("Case #%d: %lld\n",kase++,ans);
    }
    return 0;
}