标签：onclick names code 归并 esc click int empty amp
给出序列 $a_i$, 求两两之和的异或值
按位计算
计算到第 $k$ 位时，将 $a_i$ 按 $mod \ 2^{k+1}$ 后排序
当 $a_i \ mod \ 2^{k+1} + a_j \ mod \ 2^{k+1}\in [2^k, \ 2^{k+1} - 1] \cup [2^{k + 1} + 2^k, 2^{k+2}-2]$, 这两数和的第 $k$ 位为 $1$
排序时每次加入一位重排可用归并排序，找区间可以用双指针
复杂度$O(n log Max)$
#include <bits/stdc++.h> using namespace std; inline int read() { int out = 0; bool flag = false; register char cc = getchar(); while (cc < ‘0‘ || cc > ‘9‘) { if (cc == ‘-‘) flag = true; cc = getchar(); } while (cc >= ‘0‘ && cc <= ‘9‘) { out = (out << 3) + (out << 1) + (cc ^ 48); cc = getchar(); } return flag ? -out : out; } inline void write(int x) { if (x < 0) putchar(‘-‘), x = -x; if (x == 0) putchar(‘0‘); else { int num = 0; char cc[15]; while (x) cc[++num] = x % 10 + 48, x /= 10; while (num) putchar(cc[num--]); } putchar(‘\n‘); } int N, a[400010], b[400010], l1, r1, l2, r2, cnt, ans; queue<int> q1, q2; int main() { N = read(); for (int i = 1; i <= N; i++) a[i] = read(), b[i] = i; for (int k = 0; k <= 24; k++) { for (int i = 1; i <= N; i++) if (a[b[i]] & (1 << k)) q1.push(b[i]); else q2.push(b[i]); int o = 0; while (!q2.empty()) b[++o] = q2.front(), q2.pop(); while (!q1.empty()) b[++o] = q1.front(), q1.pop(); l1 = l2 = N + 1, r1 = r2 = N, cnt = 0; int Mod = (1 << (k + 1)) - 1; for (int i = 1; i <= N; i++) { o = (a[b[i]] & Mod); while (l1 >= 2 && o + (a[b[l1 - 1]] & Mod) >= (1 << k)) l1--; while (r1 >= 1 && o + (a[b[r1]] & Mod) >= (1 << (k + 1))) r1--; while (l2 >= 2 && o + (a[b[l2 - 1]] & Mod) >= (1 << (k + 1)) + (1 << k)) l2--; while (r2 >= 1 && o + (a[b[r2]] & Mod) >= (1 << (k + 2)) - 1) r2--; if (max(i + 1, l1) <= r1) cnt += r1 - max(i + 1, l1) + 1; if (max(i + 1, l2) <= r2) cnt += r2 - max(i + 1, l2) + 1; } if (cnt & 1) ans |= 1 << k; } write(ans); return 0; }
标签：onclick names code 归并 esc click int empty amp
原文地址：https://www.cnblogs.com/Urushibara-Ruka/p/13124428.html
