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