Problem D

Spiral of Numbers

Consider an infinite square grid with a clockwise spiral of consecutive positive integers. Number 1 is placed at the center, with 2 at its right, 3 bellow 2, and so one, and so forth. Having placed all number from 1 to n-1, n is placed in the same line with n-1 and n-2 unless the cell to the right of the n-1, in the [n-2, n-1] direction, is empty, in wich case n is placed in this cell. The central 11x11 square of the spiral is shown in the figure bellow.

The spiral of numbers has some intriguing features: a lot of prime numbers form diagonal lines in the spiral. This is the case of 3, 13 31, 57 and 91 in the southeast diagonal and is also the case of 5, 17 and 37 in the southwest diagonal. As you would expect this is not a general rule since 65, the next number of southwest diagonal is not a prime number.

Nevertheless the spiral is worth a little more investigation and we would like you to write a program that given a positive integer n returns its neighboring numbers in the spiral, i.e. a 3x3 square with n in the center, surrounded by the numbers that are placed in the same relative positions in the spiral.

Input

The input file contains several test cases, each of them is a positive integer less than 2^30 (a 4 byte integer) on a line by itself.

Output

For each test case, your program must write the neighboring numbers of the input number in 3 lines. Each line has 3 integers separated by semicolons. There must also be a semicolon at the end of each line.

Sample Input

11

Sample Output

9;10;27;
2;11;28;
3;12;29;

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Jos� Paulo Leal, CPUP'2003 Round 1
University of Porto Local Contest

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將 1 的位置寫成 (0,0)
因此要寫兩個轉換函數，同時計算給座標算數值，給數值算座標。

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <stdlib.h>
using namespace std;
void getXY(int n, int &x, int &y) {
    int z, w, v;
    z = sqrt(n);
    if(z%2 == 0)    z--;
    for(; z*z < n; z +=2);
    w = z/2;
    x = w, y = w;
    v = z*z;
    if(v - z < n) {
        x -= v-n;
        return;
    }
    x -= z-1, v -= z, y--;
    if(v - z+1 < n) {
        y -= v-n;
        return;
    }
    y -= z-2, v -= z-1, x++;

    if(v - z+1 < n) {
        x += v-n;
        return;
    }
    x += z-2, v -= z-1, y++;
    y += v-n;
}
int getNum(int x, int y) {
    int w = max(abs(x), abs(y));
    int z = w*2+1;
    int v = z*z, tx = w, ty = w;
    if(ty == y)
        return v - abs(tx-x);
    v -= z, tx = -w, ty = w-1;
    if(tx == x)
        return v - abs(ty-y);
    v -= z-1, ty = -w, tx = -w+1;
    if(ty == y)
        return v - abs(tx-x);
    v -= z-1, tx = w, ty = -w+1;
    if(tx == x)
        return v - abs(ty-y);
    return 0;
}
int main() {
    int n, i, j;
    while(scanf("%d", &n) == 1) {
        int x, y;
        getXY(n, x, y);
        for(i = 1; i >= -1; i--, puts(""))
            for(j = -1; j <= 1; j++)
                printf("%d;", getNum(x+j, y+i));
    }
    return 0;
}