HDU 5014 Number Sequence(xor贪心)

题意:

$给定n+1个数,a_i\in [0, n],并且a_i\neq a_j$
$现要求构造n+1个b_i,构造方式同a_i,并且使得\sum a_i\oplus b_i最大$
$输出这个sum,以及n+1个对应的b_i$

分析:

$手玩一下发现,如果n+1个数是奇数,那么0是多余的,剩余的可以两两配对$
$如果n+1是偶数直接配对即可$
$配对方式通过手玩可以发现,从大的开始,贪心补全所有的0就可以了$
$时间复杂度O(n)$

代码:

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//
// Created by TaoSama on 2016-04-25
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, a[N], mp[N];
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d", &n) == 1) {
for(int i = 0; i <= n; ++i) scanf("%d", a + i);
memset(mp, -1, sizeof mp);
if(~n & 1) mp[0] = 0;
long long ans = 0;
for(int i = n; ~i; --i) {
if(mp[i] == -1) {
int b = 32 - __builtin_clz(i);
int all = (1 << b) - 1;
mp[i] = all ^ i;
mp[all ^ i] = i;
}
ans += i ^ mp[i];
}
printf("%I64d\n", ans);
for(int i = 0; i <= n; ++i)
printf("%d%c", mp[a[i]], " \n"[i == n]);
}
return 0;
}


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