HDU 5656 CA Loves GCD (容斥)

题意：给定一个数组，每次他会从中选出若干个（至少一个数），求出所有数的GCD然后放回去，为了使自己不会无聊，会把每种不同的选法都选一遍，想知道他得到的所有GCD的和是多少。

析：枚举gcd，然后求每个gcd产生的个数，这里要使用容斥定理，f[i]表示的是 gcd 是 i 的个数，g[i] 表示的是 gcd 是 i 倍数的，f[i] = g[i] - f[j] (i|j)。

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 = 1e3 + 5;

const int maxm = 2e4 + 10;

const LL mod = 100000007;

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];

int f[maxn];

int main(){

  f[0] = 1;

  for(int i = 1; i < maxn; ++i)  f[i] = (f[i-1]<<1) % mod;

  int T;  cin >> T;

  while(T--){

    scanf("%d", &n);

    int mmax = 1;

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

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

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

      ++a[x];  mmax = max(mmax, x);

    }

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

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

        len[i] += a[j];

    LL ans = 0;

    for(int i = mmax; i; --i){

      a[i] = f[len[i]] - 1;

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

        a[i] -= a[j];

      ans = (ans + (LL)a[i] * i) % mod;

    }

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

  }

  return 0;

}

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HDU 5656 CA Loves GCD (容斥)

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