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("%d\n", tmp.count() - 1);
    }
}