題目

求$\sum_{i=1}^{n}\sum_{j=1}^n[gcd(i+j,i-j)==1]$

分析

原式=$\sum_{i=1}^{n}\sum_{j=1}^n[gcd(2i,i-j)==1]=\sum_{i=1}^{n}\sum_{j=1}^n[gcd(2i,j)==1]$

程式碼

#include <cstdio>
#define rr register
using namespace std;
const int N=20000001;
int v[N],prime[100001],cnt; long long phi[N];
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (c<48||c>57) c=getchar()
 
;
	while (c>47&&c<58) ans=(ans<<3)+(ans<<1)+c-48,c=getchar();
	return ans;
}
inline void pre(){
    for (rr int i=2;i<N;++i){
		if (!v[i]) v[i]=prime[++cnt]=i,phi[i]=i-1;
		for (rr int j=1;prime[j]*i<=N;++j){
			v[i*prime[j]]=prime[j];
			phi[i*prime[j]]=phi[i]*(prime[
 
j]-(i%prime[j]>0));
			if (i%prime[j]==0) break;
		}
	}
	for (rr int i=2;i<N;++i){
	    if (i&1) phi[i]>>=1;
	    phi[i]+=phi[i-1];
	}
}
void print(long long ans){
	if (ans>9) print(ans/10);
	putchar(ans%10+48);
}
signed main(){
	pre();
	for (rr int t=iut();t;--t)
	    print(phi[iut()]),putchar(10);
	return 0;
}