c/c++语言开发共享noi.ac#309 Mas的童年(子集乱搞)

题意

sol

记(s_i)表示前(i)个数的前缀异或和，我们每次相当于要找一个(j)满足(0 < j < i)且((s_i oplus s_j) + s_j)最大

然后下面的就和标算相差十万八千里了。

[ begin{aligned} &(s_i oplus s_j) + s_j\ =&(s_i oplus s_j oplus s_j) + ((s_i oplus s_j) & s_j )\ =&(s_i + (text{~}s_i & s_j)) end{aligned} ]

也就是对于每个(i)，我们要在前面找一个(j)使得(text{~}s[i] & s[j])最大

然后这里暴力处理子集就行了(一开始还想了半天trie树)。

加一个记忆化可以保证复杂度

最后复杂度为(o(2^{20} + n log{a_i}))

#include<bits/stdc++.h>  #define pair pair<int, int> #define mp(x, y) make_pair(x, y) #define fi first #define se second //#define int long long  #define ll long long  #define ull unsigned long long  #define fin(x) {freopen(#x".in","r",stdin);} #define fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int maxn = 3e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10; const double eps = 1e-9; template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;} template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;} template <typename a> inline void debug(a a){cout << a << 'n';} template <typename a> inline ll sqr(a x){return 1ll * x * x;} template <typename a, typename b> inline ll fp(a a, b p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;} template <typename a> a inv(a x) {return fp(x, mod - 2);} inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } int n, a[maxn], s[maxn]; bool mark[maxn]; void insert(int x) {     //if(mark[x]) return ;     mark[x] = 1;     for(int i = 0; i < 20; i++)         if((x >> i & 1) && (!mark[x ^ (1 << i)]))             insert(x ^ (1 << i)); } int query(int x) {     int ans = 0;     for(int i = 19; ~i; i--)          if((x >> i & 1) && mark[ans | 1 << i])             ans |= 1 << i;     return ans; } signed main() {     //freopen("ex_childhood2.in", "r", stdin);     n = read();     for(int i = 1; i <= n; i++) a[i] = read(), s[i] = s[i - 1] ^ a[i];     for(int i = 1; i <= n; i++) {     //  for(int j = i - 1; j >= 0; j--) chmax(ans, (s[i] ^ s[j]) + s[j]);         //for(int j = i - 1; j >= 0; j--) chmax(ans, (~s[i]) & s[j]);         int ans = query(~s[i]);         cout << s[i] + ans * 2 << ' ';          insert(s[i]);     }     puts("");     return 0; }