HihoCoder#1509 : 异或排序(二进制)

题意

题目链接

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>
#define LL long long
#define int long long  
using namespace std;
const int MAXN = 62, B = 60;
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;
LL a[MAXN], mark[MAXN];
main() {
	N = read();
	for(int i = 1; i <= N; i++) a[i] = read();
	memset(mark, -1, sizeof(mark));
	LL ans = 1ll << 60;
	for(int i = 1; i <= N - 1; i++) {
		for(int x = B; x >= 0; x--) {
			int aa = (a[i] >> x) & 1, bb = (a[i + 1] >> x) & 1;
			if(aa != bb) {
				int now = aa < bb ? 1 : 2;
				if((mark[x] != now) && (mark[x] != -1)) {puts("0"); exit(0);}
				if(mark[x] == -1) ans /= 2;
				mark[x] = now; 
				break;
			}
		}		
	}
	cout << ans;
}	
/*
*/