[NOIP 2015] 神奇的幻方

[题目链接]

          http://uoj.ac/problem/145

[算法]

         按照题意模拟即可，时间复杂度O(N^2)

[代码]

         

#include<bits/stdc++.h>
using namespace std;
#define MAXN 50

int i,j,n;
int a[MAXN][MAXN];
pair<int,int> pos[MAXN * MAXN];

template <typename T> inline void read(T &x)
{
        long long f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar())
        {
                if (c == '-') f = -f;
        }
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
}

int main() 
{
        
        read(n);
        a[1][n / 2 + 1] = 1;
        pos[1] = make_pair(1,n / 2 + 1);
        for (i = 2; i <= n * n; i++)
        {
                if (pos[i - 1].first == 1 && pos[i - 1].second != n)
                {
                        a[n][pos[i - 1].second + 1] = i;
                        pos[i] = make_pair(n,pos[i - 1].second + 1);    
                }        
                if (pos[i - 1].second == n && pos[i - 1].first != 1)
                {
                        a[pos[i - 1].first - 1][1] = i;
                        pos[i] = make_pair(pos[i - 1].first - 1,1);
                }
                if (pos[i - 1].first == 1 && pos[i - 1].second == n)
                {
                        a[2][pos[i - 1].second] = i;
                        pos[i] = make_pair(2,pos[i - 1].second); 
                } 
                if (pos[i - 1].first != 1 && pos[i - 1].second != n)
                {
                        if (!a[pos[i - 1].first - 1][pos[i - 1].second + 1])
                        {
                                a[pos[i - 1].first - 1][pos[i - 1].second + 1] = i;
                                pos[i] = make_pair(pos[i - 1].first - 1,pos[i - 1].second + 1);
                        } else
                        {
                                a[pos[i - 1].first + 1][pos[i - 1].second] = i;
                                pos[i] = make_pair(pos[i - 1].first + 1,pos[i - 1].second);
                        }
                }
        }
        for (i = 1; i <= n; i++)
        {
                for (j = 1; j <= n; j++)
                {
                        printf("%d ",a[i][j]);
                }
                printf("\n");
        }
        
        return 0;
    
}