BZOJ 2818 Gcd (莫比乌斯反演或欧拉函数)

2818: Gcd

Time Limit: 10 Sec Memory Limit:
256 MB

Submit: 2534 Solved: 1129

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Description

给定整数N，求1<=x,y<=N且Gcd(x,y)为素数的

数对(x,y)有多少对.

Input

一个整数N

Output

如题

Sample Input

Sample Output

HINT

hint

对于例子(2,2),(2,4),(3,3),(4,2)

1<=N<=10^7

Source

湖北省队互測

题目链接：

id=2818″>http://www.lydsy.com/JudgeOnline/problem.php?

id=2818

题目分析：两种姿势，莫比乌斯反演或者欧拉函数，先说简单的方法。欧拉函数，由于仅仅有一个上界n，所以变换一下1 <= x / p, y / p <= n / p，GCD(x / p, y / p) == 1，

直接求欧拉函数，令num[i]表示1到i中 1<=x，y<=i 且gcd(x,y) == 1个对数，显然有num[i] = 1 + phi[j] * 2，(1 < j <= i)，这个1指的是(1, 1)。乘2是由于(1, 2) (2, 1)算两个不同的，那么最后依据我们先前变换的公式。累加num[n / p]的值就可以

#include <cstdio>
#include <cstring>
#define ll long long
int const MAX = 1e7 + 5;
int p[MAX], phi[MAX];
bool prime[MAX];
ll num[MAX];
int pnum;void get_eular(int n)
{
    pnum = 0;
    memset(prime, true, sizeof(prime));
    for(int i = 2; i <= n; i++)
    {
        if(prime[i])
        {
            p[pnum ++] = i;
            phi[i] = i - 1;
        }
        for(int j = 0; j < pnum && i * p[j] <= n; j++)
        {
            prime[i * p[j]] = false;
            if(i % p[j] == 0)
            {
                phi[i * p[j]] = phi[i] * p[j];
                break;
            }
            phi[i * p[j]] = phi[i] * (p[j] - 1);
        }
    }
}int main()
{
    int n;
    ll ans = 0;
    scanf("%d", &n);
    get_eular(n);
    num[1] = 1;
    for(int i = 2; i <= n; i++)
        num[i] = num[i - 1] + 2 * phi[i];
    for(int i = 0; i < pnum; i++)
        if(n / p[i] > 0)
            ans += num[n / p[i]];
    printf("%lld\n", ans);
}

这题也能够用莫比乌斯反演做。还是做上述变换。1 <= x / p, y / p <= n / p，GCD(x / p, y / p) == 1，这样的题真的做烂了，懒得说了直接贴

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 1e7 + 5;
int mob[MAX], p[MAX], sum[MAX];
bool prime[MAX];
int pnum;void Mobius(int n)
{
    pnum = 0;
    memset(prime, true, sizeof(prime));
    memset(sum, 0, sizeof(sum));
    mob[1] = 1;
    sum[1] = 1;
    for(int i = 2; i <= n; i++)
    {
        if(prime[i])
        {
            p[pnum ++] = i;
            mob[i] = -1;
        }
        for(int j = 0; j < pnum && i * p[j] <= n; j++)
        {
            prime[i * p[j]] = false;
            if(i % p[j] == 0)
            {
                mob[i * p[j]] = 0;
                break;
            }
            mob[i * p[j]] = -mob[i];
        }
        sum[i] = sum[i - 1] + mob[i];
    }
}ll cal(int n)
{
    ll res = 0;
    for(int i = 1, last = 0; i <= n; i = last + 1)
    {
        last = n / (n / i);
        res += (ll) (n / i) * (n / i) * (sum[last] - sum[i - 1]);
    }
    return res;
}int main()
{
    int n;
    ll ans = 0;
    scanf("%d", &n);
    Mobius(n);
    for(int i = 0; i < pnum; i++)
        if(n / p[i] > 0)
            ans += cal(n / p[i]);
    printf("%lld\n", ans);
}