Educational Codeforces Round 7 D. Optimal Number Permutation（构造）

题意:

N≤5×105,有1∼N各出现2次的序列

如果数i在xi,yi位置,设di=yi−xi

试构造一个序列使得s=∑i=1n(n−i)⋅|di+i−n|最小

分析:

我们发现i=n的贡献一定是0,也就是说2个n可以随意放

其实可以发现一定可以构造出s=0的序列

奇数n=5：13X31 2442,缺个X填n就好了

偶数n=6:24X42 135531,缺个X填n就好了

其实规律一样的,不用分开,赛上写的急,而且也不用算n的位置,直接填到没填的地方就可以了。。。

代码:

//
//  Created by TaoSama on 2016-02-11
//  Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#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 = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, a
;

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) {
int i, j;
if(n & 1) {
for(i = 1, j = 1; j < n; ++i, j += 2)
a[i] = a[i + n - j] = j;
a[i] = a[i + n / 2 + 1] = n;
for(i = n + 2, j = 2; j < n; ++i, j += 2)
a[i] = a[i + n - j] = j;
} else {
for(i = 1, j = 2; j < n; ++i, j += 2)
a[i] = a[i + n - j] = j;
a[i] = a[i + n / 2] = n;
for(i = n + 1, j = 1; j < n; ++i, j += 2)
a[i] = a[i + n - j] = j;
}
for(int i = 1; i <= n << 1; ++i)
printf("%d%c", a[i], " \n"[i == (n << 1)]);
}
return 0;
}