[NOI2010]能量采集

Description:

求$ \sum_{i=1}^n \sum_{j=1}^m gcd(i,j)*2-1$

Hint:

\(n,m<=10^5​\)

Solution:

$ Ans=2*\sum_{i=1}^n \sum_{j=1}^m gcd(i,j) -n*m $

$\sum_{i=1}^n \sum_{j=1}^m gcd(i,j)=\sum_{T=1}^n\sum_{d=1}^{T } \mu(\frac{T}{d}) *d *\lfloor \frac{n}{T} \rfloor \lfloor \frac{m}{T} \rfloor $

$ =\sum_{T=1}^n \phi (T) *\lfloor \frac{n}{T} \rfloor \lfloor \frac{m}{T} \rfloor $

线性筛即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mxn=1e5+5;
int tot,phi[mxn],p[mxn],vis[mxn];
ll sum[mxn];

void sieve(int n)
{
    phi[1]=1;
    for(int i=2;i<=n;++i) {
        if(!vis[i]) phi[i]=i-1,p[++tot]=i;
        for(int j=1;j<=tot&&p[j]*i<=n;++j) {
            vis[p[j]*i]=1;
            if(i%p[j]) phi[p[j]*i]=phi[i]*(p[j]-1);
            else {
                phi[p[j]*i]=phi[i]*p[j];
                break ;
            }
        }
    }
    for(int i=1;i<=n;++i) sum[i]=sum[i-1]+phi[i];
} 

int main()
{
    int n,m; scanf("%d%d",&n,&m); 
    if(n>m) swap(n,m); ll ans=0; sieve(100000);
    for(int l=1,r;l<=n;l=r+1) {
        r=min(n/(n/l),m/(m/l));
        ans+=(sum[r]-sum[l-1])*(n/l)*(m/l);
    }
    printf("%lld",ans-1ll*n*m+ans); //n*m可能会炸，开long long
}

标签：lfloor,frac,int,sum,rfloor,mxn,采集,NOI2010,能量
来源： https://www.cnblogs.com/list1/p/10391082.html