codeforces 622C. Optimal Number Permutation 构造

题目链接

假设始终可以找到一种状态使得值为0， 那么两个1之间需要隔n-2个数， 两个2之间需要隔n-3个数， 两个3之间隔n-4个数。 我们发现两个三可以放到两个1之间， 同理两个5放到两个3之间....这样就构造好了。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int a[1000005];
int main()
{
    int n;
    cin>>n;
    int l = 1, r = n, cnt = 1;
    while(r>l) {
        a[l++] = cnt, a[r--] = cnt;
        cnt+=2;
    }
    l = n+1, r = l+n-2, cnt = 2;
    while(r>l) {
        a[l++] = cnt, a[r--] = cnt;
        cnt+=2;
    }
    for(int i = 1; i<=2*n; i++) {
        if(a[i])
            printf("%d ", a[i]);
        else
            printf("%d ", n);
    }
    return 0;
}