hdu 4135 容斥原理

最新推荐文章于 2019-11-13 21:09:38 发布

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本文介绍了一种算法，用于解决给定区间内与特定整数互质的数的数量问题。通过分解质因数和使用容斥原理，文章提供了两种实现方法：一种使用队列，另一种采用位运算。

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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}.

题意：

给定一个区间（a，b），求解与n其中有多少个数字和n互质

题解：

若a=1，b=13，n=6

第一步：求出n的质因子：2,3

第二步：(1,13)中是n的因子的倍数当然就不互质了(2,4,6,8,10)->n/2 6个,(3,6,9,12)->n/3 4个,(5,10)->n/5 2个

第三步：这里就需要用到容斥原理了，公式就是：n/2+n/3+n/5-n/(2*3)-n/(2*5)-n/(3*5)+n/(2*3*5).

第四步：13减去上叙式子的值

求解质因子：

void devide(LL n)//求一个数的质因子
{
    tot=0;
    for(LL i=2;i*i<=n;i++){
        if(n%i==0){
            a[tot++]=i;
            while(n%i==0)
                n=n/i;
        }
    }

    if(n>1)//这里要记得
        a[tot++]=n;
}

下面有两种方法，主要的思想都是枚举子集

方法一：队列：

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

#define LL long long
LL a[1000],tot;

void devide(LL n)//求一个数的质因子
{
    tot=0;
    for(LL i=2;i*i<=n;i++){
        if(n%i==0){
            a[tot++]=i;
            while(n%i==0)
                n=n/i;
        }
    }

    if(n>1)//这里要记得
        a[tot++]=n;
}

LL deal(LL m)//用队列数组实现容斥原理
{
    LL que[10000],pos=0,sum=0;
    que[pos++]=-1;
    for(LL i=0;i<tot;i++){
        LL k=pos;
        for(LL j=0;j<k;j++)
           que[pos++]=que[j]*a[i]*(-1);
    }
    for(LL i=1;i<pos;i++)
        sum=sum+m/que[i];
    return sum;
}

int main()
{
    int T;
    LL a,b,n;
    int cases=1;
    scanf("%d",&T);
    while(T--){
       scanf("%I64d%I64d%I64d",&a,&b,&n);
       devide(n);
       LL sum=b-deal(b)-(a-1-deal(a-1));
       printf("Case #%d: %I64d\n",cases++,sum);
    }
    return 0;
}

方法二：位运算

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

#define LL long long
LL a[1000],tot;

void devide(LL n)//求一个数的质因子
{
    tot=0;
    for(LL i=2;i*i<=n;i++){
        if(n%i==0){
            a[tot++]=i;
            while(n%i==0)
                n=n/i;
        }
    }

    if(n>1)//这里要记得
        a[tot++]=n;
}

LL deal(LL x)
{
    LL ans=0;
    for(LL i=1;i<(1LL<<tot);i++)
    {
        int sum=0,temp=1;
        for(int j=0;j<tot;j++)
        {
            if(i&(1LL<<j)){
                sum++;
                temp*=a[j];
            }
        }
        if(sum&1)
            ans+=x/temp;
        else
            ans-=x/temp;
    }
    return ans;
}
int main()
{
    int T;
    LL a,b,n;
    int cases=1;
    scanf("%d",&T);
    while(T--){
       scanf("%I64d%I64d%I64d",&a,&b,&n);
       devide(n);
       LL sum=b-deal(b)-(a-1-deal(a-1));
       printf("Case #%d: %I64d\n",cases++,sum);
    }
    return 0;
}