【SPOJ】【P5971】【LCM Sum】【题解】【数论】

传送门:http://www.spoj.com/problems/LCMSUM/

ans=n*(gx1)=n/2*((id*phi)x1+1)

g是指1..n中与n互质的数的和

Code:

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int getint(){
	int res=0;char c=getchar();
	while(!isdigit(c))c=getchar();
	while(isdigit(c))res=res*10+c-'0',c=getchar();
	return res;
}
typedef long long LL;
const int maxn=1e6+5;
bool p[maxn];
int prime[maxn];
int phi[maxn];
LL sum[maxn];
LL anss[maxn];
void getphi(){
	phi[1]=1;
	for(int i=2;i<maxn;i++){
		if(!p[i]){
			prime[++prime[0]]=i;
			phi[i]=i-1;
		}for(int j=1;i*prime[j]<maxn&&j<=prime[0];j++){
			p[i*prime[j]]=1;
			if(i%prime[j]==0){
				phi[i*prime[j]]=phi[i]*prime[j];
				break;
			}else phi[i*prime[j]]=phi[i]*(prime[j]-1);			
		}
	}for(int i=1;i<maxn;i++)sum[i]=(LL)i*phi[i];
	for(int i=1;i<maxn;i++)
	for(int j=i;j<maxn;j+=i)
	anss[j]+=sum[i];
}
int main(){
	int T=getint();
	getphi();
	while(T--){
		int n=getint();
		LL ans=(anss[n]+1)*n/2;
		printf("%lld\n",ans);
	}
	return 0;
}