HDU 1695 GCD （欧拉函数+容斥原理)

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4272    Accepted Submission(s): 1492

Problem Description

Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number pairs.
Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.

Yoiu can assume that a = c = 1 in all test cases.

 

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.
Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.

 

Output

For each test case, print the number of choices. Use the format in the example.

 

Sample Input

2 1 3 1 5 1 1 11014 1 14409 9

 

Sample Output

Case 1: 9 Case 2: 736427

Hint

For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).

 

Source

2008 “Sunline Cup” National Invitational Contest

 

Recommend

wangye

 

题意: 在1~a, 1~b中挑出(x,y)满足gcd(x,y) = k , 求(x,y) 的对数 , a,b<=10^5

思路: gcd(x, y) == k 说明x,y都能被k整除, 但是能被k整除的未必gcd=k  , 必须还要满足

互质关系. 问题就转化为了求1~a/k 和 1~b/k间互质对数的问题

可以把a设置为小的那个数, 那么以y>x来保持唯一性(题目要求, 比如[1,3] = [3,1] )

接下来份两种情况:

1. y <= a , 那么对数就是 1~a的欧拉函数的累计和(容易想到)

2. y >= a , 这个时候欧拉函数不能用了,怎么做?　　可以用容斥原理,把y与1~a互质对数问题转换为

/* ***********************************************
Author        :kuangbin
Created Time  :2013/8/19 22:08:43
File Name     :F:2013ACM练习专题学习数学HDUHDU1695GCD.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN = 10000;
int prime[MAXN+1];
void getPrime()
{
    memset(prime,0,sizeof(prime));
    for(int i = 2;i <= MAXN;i++)
    {
        if(!prime[i])prime[++prime[0]] = i;
        for(int j = 1;j <= prime[0] && prime[j] <= MAXN/i;j++)
        {
            prime[prime[j]*i] = 1;
            if(i%prime[j] == 0)break;
        }
    }
}
long long factor[100][2];
int fatCnt;
int getFactors(long long x)
{
    fatCnt = 0;
    long long tmp = x;
    for(int i = 1; prime[i] <= tmp/prime[i];i++)
    {
        factor[fatCnt][1] = 0;
        if(tmp%prime[i] == 0)
        {
            factor[fatCnt][0] = prime[i];
            while(tmp%prime[i] == 0)
            {
                factor[fatCnt][1]++;
                tmp /= prime[i];
            }
            fatCnt++;
        }
    }
    if(tmp != 1)
    {
        factor[fatCnt][0] = tmp;
        factor[fatCnt++][1] = 1;
    }
    return fatCnt;
}
int euler[100010];
void getEuler()
{
    memset(euler,0,sizeof(euler));
    euler[1] = 1;
    for(int i = 2;i <= 100000;i++)
        if(!euler[i])
            for(int j = i; j <= 100000;j += i)
            {
                if(!euler[j])
                    euler[j] = j;
                euler[j] = euler[j]/i*(i-1);
            }
}
int calc(int n,int m)//n < m,求1-n内和m互质的数的个数
{
    getFactors(m);
    int ans = 0;
    for(int i = 1;i < (1<<fatCnt);i++)
    {
        int cnt = 0;
        int tmp = 1;
        for(int j = 0;j < fatCnt;j++)
            if(i&(1<<j))
            {
                cnt++;
                tmp *= factor[j][0];
            }
        if(cnt&1)ans += n/tmp;
        else ans -= n/tmp;
    }
    return n - ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    getPrime();
    int a,b,c,d;
    int T;
    int k;
    scanf("%d",&T);
    int iCase = 0;
    getEuler();
    while(T--)
    {
        iCase++;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        if(k == 0 || k > b || k > d)
        {
            printf("Case %d: 0
",iCase);
            continue;
        }
        if(b > d)swap(b,d);
        b /= k;
        d /= k;
        long long ans = 0;
        for(int i = 1;i <= b;i++)
            ans += euler[i];
        for(int i = b+1;i <= d;i++)
            ans += calc(b,i);
        printf("Case %d: %I64d
",iCase,ans);
    }
    
    return 0;
}