SGU 109 Magic of David Copperfield II

You can change a n into n-2

if n is 2: move twice  ->  even, odd

if n is odd: move twice  ->  odd, odd

if n is even: move three times -> even, odd, odd

總之可以很容易的構造出一個方法可以從 n 變成 n-2，僅此而已

//By momo
#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
    int n; scanf("%d", &n);
    if(n & 1){
        int cnt = n, st = 1;
        for(int l = n; l > 1; l -= 2, st++){
            
            printf("%d", cnt); cnt += 2;
            for(int i = 0; i <= l; i += 2) printf(" %d %d", st+i+n*(st-1), st+i+n*(st+l-2));
            for(int i = 2; i < l-1; i += 2) printf(" %d %d", st+(st+i-1)*n, st+l-1+(st+i-1)*n);
            puts("");
            
            printf("%d", cnt); cnt += 2;
            for(int i = 1; i < l; i += 2) printf(" %d %d", st+i+n*(st-1), st+i+n*(st+l-2));
            for(int i = 1; i < l; i += 2) printf(" %d %d", st+(st+i-1)*n, st+l-1+(st+i-1)*n);
            puts("");

        }
    }
    else{
        int cnt = n+1, cnt2 = n, st = 1;
        for(int l = n; l > 2; l -= 2, st++){
            
            printf("%d", cnt2); cnt2 += 2;
            for(int i = 0; i < l; i++) printf(" %d", (i&1)? st+n*(st+i-1) : st+l-1+n*(st+i-1));
            puts("");
            
            printf("%d", cnt); cnt += 2;
            for(int i = 1; i <l-1; i++) printf(" %d", (i&1)? st+i+n*(st+l-2) : st+i+n*(st-1));
            for(int i = 0; i <  l; i++) printf(" %d", (i&1)? st+l-1+n*(st+i-1) : st+n*(st+i-1));
            puts("");

            printf("%d", cnt); cnt += 2;
            for(int i = 1; i <l-1; i++) printf(" %d", (i&1)? st+i+n*(st-1) : st+i+n*(st+l-2));
            puts("");

        }
        printf("%d %d\n%d %d %d\n", cnt2, st+n*(st-1)+1, cnt, st+n*(st-1), st+1+n*st);
    }
}