Codeforces 165 E. Compatible Numbers【子集前缀和】

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题目大意

给你一个数组，问你数组中的每个数是否可以在数组里面找到一个数和他and起来是0，如果可以就输出这个数，否则就输出-1

思路

首先很显然的是可以考虑找到每个数每一位都取反的数的子集

如果子集中存在一个数就满足，否则就不满足

然后就做一个子集前缀和

for (int i = 0; i < n; i++) {
  for (int s = 0; s <= up; s++) {
    if ((s >> i) & 1) dp[s] trans from dp[s ^ (1 << i)];
  }
}

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//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
typedef pair<int, int> pi;
typedef long long ll;
typedef double db;
#define fi first
#define se second
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
template <typename T>
void Read(T &x) {
  bool w = 1;x = 0;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') w = 0, c = getchar();
  while (isdigit(c)) {
    x = (x<<1) + (x<<3) + c -'0';
    c = getchar();
  }
  if (!w) x = -x;
}
template <typename T>
void Write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) Write(x / 10);
  putchar(x % 10 + '0');
}
//----------------------------------------------
const int N = (1 << 24) + 10;
int n, dp[N], a[N];
int bitlen(int x) {
  int res = 0;
  while (x) ++res, x >>= 1;
  return max(1, res);
}
int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif
  memset(dp, -1, sizeof(dp));
  Read(n);
  int len = 0;
  fu(i, 1, n) {
    Read(a[i]);
    dp[a[i]] = a[i];
    len = max(len, bitlen(a[i]));
  }
  int up = (1 << len) - 1;
  fu(i, 0, len)
    fu(s, 0, up) if (dp[s] == -1 && ((s >> i) & 1))
      dp[s] = dp[s ^ (1 << i)];
  fu(i, 1, n) Write(dp[up ^ a[i]]), putchar(' ');
  return 0;
}