SPOJ 7001. Visible Lattice Points 【mobius反演】

最新推荐文章于 2019-12-23 16:58:43 发布

zz_1215

最新推荐文章于 2019-12-23 16:58:43 发布

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本文链接：https://blog.youkuaiyun.com/zz_1215/article/details/40659431

本文介绍了一种高效的素数筛法，并利用数论中的莫比乌斯函数解决特定数学问题。通过初始化素数数组和莫比乌斯函数数组，实现对指定范围内整数性质的有效计算。

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#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;
#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define foreach(c,itr)  for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define pf              push_front
#define X               first
#define Y               second
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MC(a,b)			memcpy(a,b,sizeof(b))
#define MAX(a,b)        ((a)>(b)?(a):(b))
#define MIN(a,b)        ((a)<(b)?(a):(b))
#define read            freopen("out.txt","r",stdin)
#define write           freopen("out2.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-6;
const double pi = acos(-1.0);
const int maxn = 1000001;

int prime[maxn];
int mu[maxn];
int smu[maxn];
bool check[maxn];
int tot;
int n;

void init(){
	mu[1] = 1;
	tot = 0;
	for (int now = 2; now < maxn; now++){
		if (!check[now]){
			prime[tot++] = now;
			mu[now] = -1;
		}
		for (int i = 0; i < tot; i++){
			if (now*prime[i]>maxn) break;
			check[now*prime[i]] = true;
			if (now%prime[i] == 0){
				mu[now*prime[i]] = 0;
				break;
			}
			else{
				mu[now*prime[i]] = -mu[now];
			}
		}
	}
	for (int i = 1; i < maxn; i++){
		smu[i] = smu[i - 1] + mu[i];
	}
}


int main(){
	init();
	int T;
	cin >> T;
	while (T--){	
		cin >> n;
		i64 ans = 3;
		i64 ans2 = 0;
		i64 ans3 = 0;
		int temp;
		i64 t2;
		for (int i = 1; i <= n; i=temp+1){
			temp = (n / (n / i));
			t2 = n / i;
			t2 *= t2;
			ans2 += t2*(smu[temp] - smu[i - 1]);
			ans3 += t2*(i64)(n / i)*(smu[temp]-smu[i-1]);
		}
		ans2 *= 3;
		ans += ans2 + ans3;
		cout << ans << endl;
	}		
	return 0;
}