c/c++语言开发共享HihoCoder#1509 : 异或排序(二进制)

题意

sol

挺简单的吧。考虑两个元素什么时候不满足条件

设(a_i)与(a_i + 1)最高的不同位分别为0 1，显然(s)的这一位必须为(0)，否则这一位必须为(1)

剩下的就没有限制条件了

时间复杂度：(nlogn)??????!!!!!!

#include<bits/stdc++.h> using namespace std; const int maxn = 30001; inline int read() {     int x = 0, f = 1; char c = getchar();     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[5][maxn], rak[5][maxn], base = 1; bitset<maxn> b[5][175]; main() {     n = read(); base = sqrt(n);     for (int i = 1; i <= n; i++) for (int j = 0; j < 5; j++) a[j][i] = read(), rak[j][a[j][i]] = i;     for (int j = 0; j < 5; j++)          for (int i = 1; i * base <= n; i++)              for(int k = 1; k <= i * base; k++) b[j][i].set(rak[j][k]);       for (int i = 1; i <= n; i++) {         bitset<maxn> tmp, cal; tmp.set();         for(int j = 0; j < 5; j++) {             cal.reset();             int now = a[j][i] / base;             cal |= b[j][now];             for(int k = now * base + 1; k <= a[j][i]; k++) cal.set(rak[j][k]);              tmp &= cal;         }         printf("%dn", tmp.count() - 1);     } }