Submission #71436263

Source Code Expand

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#include <stdio.h>
int main(void) {
  int N;
  int i, j;
  if (scanf("%d", &N) != 1) return 1;
  for (i = 0; i < N; i++) {
    for (j = 0; j < N; j++) {
      printf(" %d" + !j, (i * N + j + 1) ^ (N % 2 == 1 && i == N / 2 && (j == N / 2 || j == N / 2 - 1)));
    }
    putchar('\n');
  }
  return 0;
}
/*
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הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
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#include <stdio.h>

int main(void) {
	int N;
	int i, j;
	if (scanf("%d", &N) != 1) return 1;
	for (i = 0; i < N; i++) {
		for (j = 0; j < N; j++) {
			printf(" %d" + !j, (i * N + j + 1) ^ (N % 2 == 1 && i == N / 2 && (j == N / 2 || j == N / 2 - 1)));
		}
		putchar('\n');
	}
	return 0;
}

/*

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みたいに、単純に順番に並べる
行：1個下に行くと N**2 増える
列：1個右に行くと N 増える

最初の行の合計は N*(N+1)/2
最後の列の合計は 最初の行の N 倍 ＝ N*N*(N+1)/2
よって、最後の列は、最初の行から (N-1)*N*(N+1)/2 増えている

ぶつからないか調べる

def check(n):
    deltas = [n*n*k for k in range(n)] + [(n-1)*n*(n+1)//2-n*k for k in range(n)]
    return len(set(deltas)) == n * 2

all([check(n) for n in range(2, 51)])

→ False

all([check(n) for n in range(2, 51, 2)])

→ True

よって、N が偶数のときは、これでおｋ

def check2(n):
    deltas = [n*n*k for k in range(n)] + [(n-1)*n*(n+1)//2-n*k for k in range(n)]
    return len(set(deltas)) == n * 2 - 1

all([check2(n) for n in range(3, 51, 2)])

→ True

よって、N が奇数のとき、必ずちょうど1個だけ被っている
→ 多分真ん中だろう
→ 真ん中付近の要素を入れ替え、列の和を±1して、被らなくする

*/

Submission Info

Judge Result