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): 2346    Accepted Submission(s): 850

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).
 

          

题目大意：

给你a, b, c, d, k; 找出这样的一队 x, y， 使得 gcd(x , y) = k, 并且x ∈[1, b], y ∈ [1, d], 问有多少对符合要求的(x, y)。

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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y的质数因子在1~a范围内能整除的个数(质数分解和容斥关系)，dfs一下即可。

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1695

代码：

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <vector>
using namespace std;

const int N = 100005;
typedef long long LL;

LL prime[N];
LL phi[N];
bool is_prime[N];
vector<LL> link[N];

void get_phi()  //筛法欧拉函数
{
    LL i, j, k = 0;
    phi[1] = 1;
    for(i = 2; i < N; i++)
    {
        if(is_prime[i] == false)
        {
            prime[k++] = i;
            phi[i] = i-1;
        }
        for(j = 0; j<k && i*prime[j]<N; j++)
        {
            is_prime[ i*prime[j] ] = true;
            if(i%prime[j] == 0)
            {
                phi[ i*prime[j] ] = phi[i] * prime[j];
                break;
            }
            else
            {
                phi[ i*prime[j] ] = phi[i] * (prime[j]-1);
            }
        }
    }
}

void init()     //求每一个数的质因数，vector储存
{
    LL i, j, k;
    for(i = 1; i < N; i++)
    {
        k = i;
        for(j = 0; prime[j]*prime[j] <= k; j++)
        {
            if(k%prime[j] == 0)
            {
                link[i].push_back(prime[j]);
                while(k%prime[j] == 0)
                    k /= prime[j];
            }
            if(k == 1) break;
        }
        if(k > 1) link[i].push_back(k);
    }
}

LL dfs(LL a, LL b, LL cur)  //容斥原理计算
{
    LL i, k, res = 0;
    for(i = a; i < link[cur].size(); i++)
    {
        k = b/link[cur][i];
        res += k - dfs(i+1, k, cur);
    }
    return res;
}

int main()
{
    LL i, a, b, c, d, k, sum, t, zz = 1;
    get_phi();
    init();
    scanf("%I64d", &t);
    while(t--)
    {
        scanf("%I64d %I64d %I64d %I64d %I64d", &a, &b, &c, &d, &k);
        if(k == 0 || k > b || k > d)
        {
            printf("Case %I64d: 0\n", zz++);
            continue;
        }
        if(b > d) swap(b, d);
        b /= k;
        d /= k;
        sum = 0;
        for(i = 1; i <= b; i++)
        {
            sum += phi[i];
        }
        for(i = b+1; i <= d; i++)
        {
            sum += b - dfs(0, b, i);
        }
        printf("Case %I64d: %I64d\n", zz++, sum);
    }

    return 0;
}