BZOJ4804: 欧拉心算

Time Limit: 15 Sec Memory Limit: 256 MB
Submit: 434 Solved: 262
[Submit][Status][Discuss]

Description

给出一个数字N

Input

第一行为一个正整数T，表示数据组数。
接下来T行为询问，每行包含一个正整数N。
T<=5000,N<=10^7

Output

按读入顺序输出答案。

Sample Input

1

10

Sample Output

136

HINT

Source

By FancyCoder

题解

\[\sum_{i=1}^n \sum_{j=1}^n \phi(gcd(i,j)) \]

\[=\sum_{i=1}^n \sum_{j=1}^n \sum_{d=1}^n \phi(d)[gcd(i,j) =d] \]

\[=\sum_{d=1}^n\phi(d)\sum_{i=1}^n \sum_{j=1}^n[gcd(i,j)=d] \]

\[=\sum_{d=1}^n\phi(d)\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{n}{d} \rfloor}[gcd(i,j)=1] \]

不难发现\(\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{n}{d} \rfloor}[gcd(i,j)=1]\)其实就是两倍欧拉函数前缀和减去1（因为(1,1)和(1,1)重复）

于是线性筛\(\phi\)，求前缀和即可

好像\(n,n\)改成\(n,m\)也可做，不过线筛起来挺（我）麻（不）烦（会）

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
inline long long  max(long long a, long long b){return a > b ? a : b;}
inline long long  min(long long a, long long b){return a < b ? a : b;}
inline void swap(long long &x, long long &y){long long  tmp = x;x = y;y = tmp;}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}
const long long INF = 0x3f3f3f3f;
const long long MAXN = 10000000;

long long p[MAXN + 1], bp[MAXN + 1], tot, phi[MAXN + 1];
void make_phi()
{
	phi[1] = 1;
	for(long long i = 2;i <= MAXN;++ i)
	{
		if(!bp[i]) phi[i] = i - 1, p[++ tot] = i;
		for(long long j = 1;j <= tot && i * p[j] <= MAXN;++ j)
		{
			bp[i * p[j]] = 1;
			if(i % p[j] == 0)
			{
				phi[i * p[j]] = phi[i] * p[j];
				break;
			}
			phi[i * p[j]] = phi[i] * (p[j] - 1);
		}
	}
	for(long long i = 1;i <= MAXN;++ i) phi[i] += phi[i - 1]; 
} 
long long t, n;
int main()
{
	make_phi();
	read(t);
	for(;t;--t)
	{
		read(n);
		long long ans = 0, r;
		for(long long d = 1;d <= n;++ d)
		{
			r = min(n/(n/d), n);
			ans += (phi[r] - phi[d - 1]) * ((phi[n/d] << 1) - 1);
			d = r;
		}
		printf("%lld\n", ans);
	}
 	return 0;
}