HDU 5212 Code (莫比乌斯反演)

题意：给定上一个数组，求

析：

其中，f(d)表示的是gcd==d的个数，然后用莫比乌斯反演即可求得，len[i]表示能整队 i 的个数，可以线性筛选得到，

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

//#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e4 + 5;

const int maxm = 2e4 + 10;

const LL mod = 10007;

const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};

const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

int a[maxn], len[maxn];

bool vis[maxn];

int prime[maxn], mu[maxn];

void Moblus(){

  mu[1] = 1;

  int tot = 0;

  for(int i = 2; i < maxn; ++i){

    if(!vis[i])  prime[tot++] = i, mu[i] = -1;

    for(int j = 0; j < tot; ++j){

      int t = i * prime[j];

      if(t >= maxn)  break;

      vis[t] = 1;

      if(i % prime[j] == 0)  break;

      mu[t] = -mu[i];

    }

  }

}

int f[maxn];

int main(){

  Moblus();

  while(scanf("%d", &n) == 1){

    ms(a, 0);  ms(len, 0);  ms(f, 0);

    for(int i = 0; i < n; ++i){

      int x;  scanf("%d", &x);

      ++a[x];

    }

    for(int i = 1; i < maxn; ++i)

      for(int j = i; j < maxn; j += i)

        len[i] += a[j];

    for(int i = 1; i < maxn; ++i)

      for(int j = i, cnt = 1; j < maxn; j += i, ++cnt)

        f[i] += mu[cnt] * sqr(len[j]);

    LL ans = 0;

    for(int i = 2; i < maxn; ++i)

      ans = (ans + i * (i-1) * (LL)f[i]) % mod;

    printf("%lld\n", ans);

  }

  return 0;

}