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GCD - Extreme (II) UVA - 11426 尤拉函式_數學推導

Code:

#include<cstdio>              
using namespace std;
const int maxn=4000005;
const int R=4000002;
const int N=4000002;
long long sumv[maxn],f[maxn];
int phi[maxn],prime[maxn],vis[maxn];
void solve(){
	phi[1]=1;
	int cnt=0;
	for(int i=2;i<=R;++i){
		if(!vis[i])prime[++cnt]=i,phi[i]=i-1;
		for(int
j=1;j<=cnt&&prime[j]*i<=R;++j){ vis[prime[j]*i]=1; if(i%prime[j]!=0)phi[i*prime[j]]=phi[i]*(prime[j]-1); else{ phi[i*prime[j]]=phi[i]*prime[j]; break; } } } for(int i=1;i<N;++i) for(int j=i*2;j<N;j+=i) f[j]+=i*phi[j/i]; for(int i=1;i<N;++i) sumv[
i]=sumv[i-1]+f[i]; } int main(){ solve(); while(1){ int n;scanf("%d",&n); if(!n)break; printf("%lld\n",sumv[n]); } return 0; }