BZOJ3944 Sum

@(BZOJ)[杜教筛]

Description

Input

一共T+1行
第1行为数据组数T(T<=10)
第2~T+1行每行一个非负整数N，代表一组询问

Output

一共T行，每行两个用空格分隔的数ans1,ans2

Sample Input

6
1
2
8
13
30
2333

Sample Output

1 1
2 0
22 -2
58 -3
278 -3
1655470 2

Solution

入门模板题.
注意一点点常数优化的技巧: 调用函数时传入参数的速度是很慢的, 要尽量控制传入参数的数量和参数的类型大小.

#include <cstdio>
#include <cstring>

const long long BND = 4000000;
int n;
long long sum1[BND], sum2[BND], vis[600], _sum[600];

inline void pretreat()
{
	static int vis[BND], prm[BND], phi[BND], mu[BND];
	memset(phi, 0, sizeof(phi));
	int cnt = 0;
	mu[1] = phi[1] = 1;

	for(int i = 2; i < BND; ++ i)
	{
		if(! phi[i])
			prm[cnt ++] = i, mu[i] = -1, phi[i] = i - 1;

		for(int j = 0; j < cnt && i * prm[j] < BND; ++ j)
		{
			int crt = i * prm[j];

			if(! (i % prm[j]))
			{
				mu[crt] = 0;
				phi[crt] = phi[i] * prm[j]; // phi[k * prm[j] ^ (k + 1)] / phi[k * prm[j] ^ k] = prm[j]
				break;
			}

			mu[crt] = - mu[i];
			phi[crt] = phi[prm[j]] * phi[i];
		}
	}
	
	sum1[0] = sum2[0] = 0;

	for(int i = 1; i < BND; ++ i)
		sum1[i] = sum1[i - 1] + phi[i], sum2[i] = sum2[i - 1] + mu[i];
}

long long getPhiSum(int crt)
{
	if(crt < BND)
		return sum1[crt];

	int tmp = n / crt;

	if(vis[tmp])
		return _sum[tmp];

	_sum[tmp] = crt * ((long long)crt + 1) / 2;

	for(int i = 2, lst; i <= crt; i = lst + 1) {
		lst = crt / (crt / i), _sum[tmp] -= getPhiSum(crt / i) * (lst - i + 1);
		if (lst==n) break;
	}

	vis[tmp] = 1;
	return _sum[tmp];
}

long long getMuSum(int crt)
{
	if(crt < BND)
		return sum2[crt];

	int tmp = n / crt;

	if(vis[tmp])
		return _sum[tmp];

	_sum[tmp] = 1;

	for(int i = 2, lst; i <= crt; i = lst + 1) {
		lst = crt / (crt / i), _sum[tmp] -= getMuSum(crt / i) * (lst - i + 1);
		if (lst==n) break;
	}
	
	vis[tmp] = 1;
	return _sum[tmp];
}

int main()   
{
	#ifndef ONLINE_JUDGE
	freopen("BZOJ3944.in", "r", stdin);
	freopen("BZOJ3944.out", "w", stdout);
	#endif
	
	pretreat();
	int T;

	for(scanf("%d", &T); T --;)
	{   
		scanf("%d", &n);
		memset(vis, 0, sizeof(vis));
		long long ans1 = getPhiSum(n);
		memset(vis, 0, sizeof(vis));
		long long ans2 = getMuSum(n);
		printf("%lld %lld\n", ans1, ans2);
	}
}