脚趾踢到钢板，撞到硬题了，要求有点古怪，期待你进来
本以为很简单的算法题，幻想几分钟几行代码搞定，没想到几年都没搞定，踢到钢板了
各位神人果断出手，救题于千军万马中，招数任出
当然，能够不用条件句那就最好，能不借助数组就不借助数组，当然，不是说不可以
输入输出行数，得出下图数字阵列
来个java版：
public class Test {
public static int N = 10;
public static int value(int px, int py, int di, int dj, int s, int n) {
for (int i = 0; i < n; i++) {
if (px == i + di && py == i + dj) {
return s + i + 1;
}
if (i != 0) {
if (px == i + di && py == 0 + dj) {
return s + 3 * n - 2 - i;
}
if (px == n - 1 + di && py == i + dj) {
return s + 2 * n - 1 - i;
}
}
}
if (n > 3) {
return value(px, py, 2 + di, 1 + dj, s + 3 * n - 3, n - 3);
}
throw new RuntimeException("something goes wrong.");
}
public static void main(String[] args) throws Exception {
int width = Integer.toString((int) ((N * N - N) / 2 + N)).length();
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (i >= j) {
System.out.print(String.format("%" + width + "s", value(i, j, 0, 0, 0, N)) + "\t");
}
}
System.out.println();
}
}
}
public class ElevenTest{
public static void main(String[] args){
System.out.println("1");
System.out.println("24 2");
System.out.println("23 25 3");
System.out.println("22 39 26 4");
System.out.println("21 38 40 27 5 ");
System.out.println("20 37 45 41 28 6");
System.out.println("19 36 44 43 42 29 7");
System.out.println("18 35 34 33 32 31 30 8 ");
System.out.println("17 16 15 14 13 12 11 10 9 ");
}
}
吼吼吼，最简单的做法
如果有跟我一样有点小白，又想知道如何解，就留个言，有高人解答了，我们就可以知道怎么做咯
插眼
暂时没有思路的，可以先关注本题哈，免得退出就再也找不到进来的门了
已解决
package com.eleven;
/**
* @author zhaojinhui
* @date 2021/6/4 15:11
* @apiNote
*/
public class ElevenTest {
private static int xOffset = 0;
private static int yOffset = 0;
private static int max = 0;
private static int lastOutIndex = 0;
public static void main(String[] args) {
int n = 9;
int[][] arr = new int[n][n];
int sub = n;
lastOutIndex = n -1;
while (sub > 1){
//System.out.println("max == : " + max);
test(sub,arr);
sub = sub - 3;
xOffset++;
yOffset = yOffset + 2;
lastOutIndex--;
}
showArr(arr);
}
private static void test(int n,int[][] arr){
int doubleN = 2 * (n + max);
//System.out.println("此轮中间数为 + " + (n + max));
//System.out.println("当前的lastIndex：" + lastOutIndex);
//String format = "arr[%d][%d] = %d";
for (int i = 0; i < n; i++) {
int outIndex = i + yOffset;
int inIndex = i + xOffset;
arr[i + yOffset][i + xOffset] = i + 1 + max;
//System.out.println(String.format(format,outIndex,inIndex,arr[outIndex][inIndex]));
arr[lastOutIndex][i + xOffset] = doubleN - arr[i + yOffset][i + xOffset];
//System.out.println(String.format(format,lastOutIndex,inIndex,arr[lastOutIndex][inIndex]));
}
for (int i = n - 2; i >= 1; i--) {
arr[i + yOffset][0 + xOffset] = arr[i + 1 + yOffset][0 + xOffset] + 1;
//System.out.println(String.format(format, i + yOffset,0 + xOffset,arr[i + yOffset][0 + xOffset]));
if(i == 1){
max = arr[i + yOffset][0 + xOffset];
//System.out.println("max == : " + max);
}
}
}
private static void showArr(int[][] arr){
for (int[] ints : arr) {
for (int anInt : ints) {
System.out.print(anInt == 0 ? " \t" : anInt + "\t");
}
System.out.println();
}
}
}
运行结果
1
24 2
23 25 3
22 39 26 4
21 38 40 27 5
20 37 45 41 28 6
19 36 44 43 42 29 7
18 35 34 33 32 31 30 8
17 16 15 14 13 12 11 10 9
留个言，我也期待问题的答案
N = 9
def value(px, py, di=0, dj=0, s=0, n=N):
for i in range(n):
if px == i + di and py == i + dj:
return s + i + 1
if i != 0:
if px == i + di and py == 0 + dj:
return s + 3 * n - 2 - i
if px == n - 1 + di and py == i + dj:
return s + 2 * n - 1 - i
if n > 3:
return value(px, py, 2 + di, 1 + dj, s + 3 * n - 3, n - 3)
width=len(str(int((N*N-N)/2+N)))
for i in range(N):
for j in range(N):
if i >= j:
print(str(value(i,j)).rjust(width), end='\t')
print()
1
24 2
23 25 3
22 39 26 4
21 38 40 27 5
20 37 45 41 28 6
19 36 44 43 42 29 7
18 35 34 33 32 31 30 8
17 16 15 14 13 12 11 10 9
为了研究楼上的代码，还特定去苗了一下python语法
/*
1
15 2
14 16 3
13 21 17 4
12 20 19 18 5
11 10 9 8 7 6
*/
class Pos {
x: number;
y: number;
constructor(x: number, y: number) {
this.x = x;
this.y = y;
}
CanMove(dir: Dir, xMax: number, yMax: number): boolean {
var nexty = dir.yinc + this.y;
var nextx = dir.xinc + this.x;
return !(nextx < 0 || nexty < 0 || nextx >= xMax + dir.xinc || nexty >= yMax);
}
Move(dir: Dir): void {
this.x += dir.xinc;
this.y += dir.yinc;
}
public toString = (): string => {
return `(x: ${this.x} y: ${this.y})`;
}
}
class Dir {
xinc: number;
yinc: number;
constructor(x: number, y: number) {
this.xinc = x;
this.yinc = y;
}
public toString = (): string => {
return `(xinc: ${this.xinc} yinc: ${this.yinc})`;
}
}
function NextDir(current: Dir): Dir {
if (current.xinc == 1 && current.yinc == 1)
return new Dir(-1, 0);
if (current.xinc == -1 && current.yinc == 0)
return new Dir(0, -1);
if (current.xinc == 0 && current.yinc == -1)
return new Dir(1, 1);
}
function Repeat(str: string, n: number): string[] {
var arr = [];
for (var i = 0; i < n; i++) {
arr.push(str);
}
return arr;
}
function CircleNum(n: number): string {
var arr = [];
for (let i = 0; i < n; i++) {
var ele = Repeat("0", i + 1);
arr.push(ele);
}
var currentPos = new Pos(0, 0);
var currentDir = new Dir(1, 1);
var xmax = 1;
var ymax = n;
var val = 1;
while (arr[currentPos.y][currentPos.x] == "0") {
arr[currentPos.y][currentPos.x] = val++;
if (currentPos.CanMove(currentDir, xmax, ymax)) {
if (arr[currentPos.y + currentDir.yinc][currentPos.x + currentDir.xinc] != "0")
currentDir = NextDir(currentDir);
currentPos.Move(currentDir);
}
else {
currentDir = NextDir(currentDir);
currentPos.Move(currentDir);
}
xmax = arr[currentPos.y].length;
if (arr[currentPos.y][currentPos.x] != "0")
break;
}
return arr.map(x => x.join(" ")).join("\n");
}
console.log(CircleNum(6));
typescript
游戏写多了，按平时写游戏的方法定义了pos和dir
我导师给的答案出来了，不需要任何条件语句，比如if或?:等，至于什么原理，我就不知道了，我导师说这是星辰运转之理，不需要人为的条件处理
static int ggg(int A1) { return ((1 - A1 - Math.Abs(A1)) + Math.Abs((1 - A1 - Math.Abs(A1))) - 2) / -2; }
static int hhh(int u,int n) { return Convert.ToInt32((u * 6 + 3 - n * 9) / 2.0 * n); }
static void twine(int u)
{
for (int i = 0; i < u; i++)
{
int x1 =u- i;
int x2 = i*2-u+1 ;
int ei = hhh(u,(u-1-i));
for (int j = 0; j <= i; j++)
{
int y1 = j * 2 + 1;
int y2 = u-1- j;
int g1 = ggg(i + 1 - y1) * ggg(y2 + 1 - i);
int g2= ggg(j + 1 - x1) * ggg(x2 + 1 - j);
int n = g1 * (hhh(u,j) + (y2 - y1 + 1) * 2 + (y2 - i));
n += g2 *( ei + (x2 - x1 + 1) + (x2 - j))+(1 - g1 - g2) * (hhh(u, (i - j)) + i - (i - j) * 2);
Console.Write("{0,4}", n+1);
}
Console.WriteLine();
}
}
运行效果如下：
各位神牛能解释一下是怎么实现的吗？