2005: [Noi2010]能量采集

2005: [Noi2010]能量采集

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分析

答案要求 $ans=2sumlimits_{x=1}^n sumlimits_{y=1}^mgcd(x,y)-n imes m$

主要求出中间的那一块就好了。

思路一：

　　容斥：f[i]表示gcd(x,y)=i的对数。(n/i)*(m/i)是gcd(x,y)=i的倍数 的对数。那么f[i] -= f[2*i] - f[3*i] ...就好了。

思路二：

　　莫比乌斯反演

$ sumlimits_{x=1}^n sumlimits_{y=1}^mgcd(x,y)$
$=sumlimits_p^{min(n,m)} p sumlimits_{x=1}^n sumlimits_{y=1}^m[gcd(x,y)=p]$
$=sumlimits_p^{min(n,m)} p sumlimits_d^{min(frac{n}{p},frac{m}{p})}μ(d)frac{n}{pd} imes frac{m}{pd}$
$ sumlimits_T^{min(n,m)} frac{n}{T} imes frac{m}{T} sumlimits_{p|T}p imes μ(frac{T}{p})$
$=sumlimits_T^{min(n,m)} frac{n}{T} imes frac{m}{T} φ(T)$

容斥：

#include<cstdio>
#include<iostream>
 
using namespace std;
 
typedef long long LL;
const int N = 100100;
LL f[N];
 
int main() {
    LL n,m,ans = 0;
    cin >> n >> m;
    int mi = min(n,m);
    for (int i=mi; i>=1; --i) {
        f[i] = (n / i) * (m / i);
        for (int j=i+i; j<=mi; j+=i) {
            f[i] -= f[j];
        }
        ans += f[i] * i;
    }
    cout << 2 * ans - n * m;
    return 0;
}

莫比乌斯+欧拉函数：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 100010;
int prime[N],phi[N],tot;
bool noprime[N];

void getphi(int n) {
    phi[1] = 1;
    for (int i=2; i<=n; ++i) {
        if (!noprime[i]) prime[++tot] = i,phi[i] = i-1;
        for (int j=1; j<=tot&&prime[j]*i<=n; ++j) {
            noprime[i * prime[j]] = true;
            if (i % prime[j] == 0) {
                phi[i * prime[j]] = phi[i] * prime[j];
                break; 
            }
            phi[i * prime[j]] = phi[i] * (prime[j]-1); 
        }
    }
    for (int i=1; i<=n; ++i) phi[i] += phi[i-1];
}
int main() {
    
    LL n,m;
    cin >> n >> m;
    LL mi = min(n,m);
    getphi((int)mi);
    LL ans = 0,pos = 0;
    for (int T=1; T<=mi; T=pos+1) {
        pos = min(n/(n/T),m/(m/T));
        ans += (n/T) * (m/T) * (phi[pos] - phi[T-1]);
    }
    cout << ans * 2 - n * m;    
    return 0;
}