HDU1695: GCD 题解

最新推荐文章于 2024-11-07 08:22:46 发布

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#莫比乌斯反演

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本文深入探讨了数学算法中关于最大公约数(gcd)的求解技巧，通过数学归纳法和莫比乌斯反演公式推导出高效计算无序数对的方法，并提供了完整的C++代码实现。

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\sum i = 1 N \sum j = 1 M [g c d (i, j) = k] = \sum i = 1 ⌊ N k ⌋ \sum j = 1 ⌊ M k ⌋ [g c d (i, j) = 1]

$\sum_{i=1}^{N}\sum_{j=1}^{M}[gcd(i,j)=k]=\sum_{i=1}^{\left\lfloor\frac{N}{k}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac{M}{k}\right\rfloor}[gcd(i,j)=1]$

f (k) = \sum i = 1 ⌊ N k ⌋ \sum j = 1 ⌊ M k ⌋ [g c d (i, j) = k]

$f(k)=\sum_{i=1}^{\left\lfloor\frac{N}{k}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac{M}{k}\right\rfloor}[gcd(i,j)=k]$

令 n = ⌊ N k ⌋, m = ⌊ M k ⌋

$令n=\left\lfloor\frac{N}{k}\right\rfloor,m=\left\lfloor\frac{M}{k}\right\rfloor$

F (k) = \sum n | d f (d) = ⌊ n k ⌋ ⌊ m k ⌋

$F(k)=\sum_{n\left.\right|d}f(d)=\left\lfloor\frac{n}{k}\right\rfloor\left\lfloor\frac{m}{k}\right\rfloor$
*Mobius Inversion

f (n) = \sum n | d F (d) μ (d n)

$f(n)=\sum_{n\left.\right|d}F(d)\mu(\frac{d}{n})$

f (1) = \sum i = 1 m i n (n, m) F (i) μ (i) = \sum i = 1 m i n (n, m) ⌊ n i ⌋ ⌊ m i ⌋ μ (i)

$f(1)=\sum_{i=1}^{min(n,m)}F(i)\mu(i)=\sum_{i=1}^{min(n,m)}\left\lfloor\frac{n}{i}\right\rfloor\left\lfloor\frac{m}{i}\right\rfloor\mu(i)$
这个分块搞一下就好了
注意几个细节和坑点：
1. k可能等于0；刚开始要保证

N<MN<M $N<M$
2. 题目要求的是无序数对，所以先对N,M做一遍记为ans1，再对N,N做一遍记为ans2,ans2/2就是重复的部分，ans=ans1-ans2/2

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <utility>
#include <cctype>
#include <algorithm>
#include <bitset>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <cmath>
#define LL long long
#define LB long double
#define x first
#define y second
#define Pair pair<int,int>
#define pb push_back
#define pf push_front
#define mp make_pair
#define LOWBIT(x) x & (-x)
using namespace std;

const int MOD=1e9+9;
const LL LINF=2e16;
const int INF=2e9;
const int magic=348;
const double eps=1e-10;
const double pi=3.14159265;

inline int getint()
{
    char ch;int res;bool f;
    while (!isdigit(ch=getchar()) && ch!='-') {}
    if (ch=='-') f=false,res=0; else f=true,res=ch-'0';
    while (isdigit(ch=getchar())) res=res*10+ch-'0';
    return f?res:-res;
}

int N,M,k,n,m;
int mu[100048],sum[100048];
bool isprime[100048];int prime[100048],tot=0;

inline void init_mu()
{
    int i,j;
    mu[1]=sum[1]=1;memset(isprime,true,sizeof(isprime));
    for (i=2;i<=100000;i++)
    {
        if (isprime[i]) prime[++tot]=i,mu[i]=-1;
        sum[i]=sum[i-1]+mu[i];
        for (j=1;j<=tot && (long long)(i)*prime[j]<=100000;j++)
        {
            isprime[i*prime[j]]=false;
            if (i%prime[j]==0) mu[i*prime[j]]=0; else mu[i*prime[j]]=-mu[i];
        }
    }
}

int main ()
{
    int ca,x,pos,lastpos;ca=getint();
    init_mu();LL ans1,ans2;int tt=0;
    while (ca--)
    {
        N=getint();N=getint();M=getint();M=getint();k=getint();
        if (k==0)
        {
            printf("Case %d: 0\n",++tt);
            continue;
        }
        if (N>M) swap(N,M);
        n=N/k;m=M/k;lastpos=1;pos=0;ans1=0;ans2=0;
        while (pos<=n && lastpos<=n)
        {
            pos=min(n/(n/lastpos),m/(m/lastpos));if (pos>n) break;
            ans1+=(long long)(sum[pos]-sum[lastpos-1])*(long long)(n/pos)*(m/pos);
            lastpos=pos+1;
        }
        lastpos=1;pos=0;
        while (pos<=n && lastpos<=n)
        {
            pos=n/(n/lastpos);if (pos>n) break;
            ans2+=(long long)(sum[pos]-sum[lastpos-1])*(long long)(n/pos)*(n/pos);
            lastpos=pos+1;
        }
        printf("Case %d: %lld\n",++tt,ans1-(ans2>>1));
    }
    return 0;
}