题目：
一、编写一个程序，输入一个数字n，让其组成一个二维数组int i[][]=i[n][n]。然后输出这个数组的所有值，并让这些值呈螺旋状由外到里排列出来。

如：输入一个5，则数组为 int i[][]=i[5][5]
输出结果为：

i[0][0] i[0][1] i[0][2] i[0][3] i[0][4]
i[3][0] i[3][1] i[3][2] i[3][3] i[1][0]
i[2][4] i[4][3] i[4][4] i[3][4] i[1][1]
i[2][3] i[4][2] i[4][1] i[4][0] i[1][2]
i[2][2] i[2][1] i[2][0] i[1][4] i[1][3]

即——若i[0][0]=1则最终显示结果为：

1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
------------------------------------------
从1到N(100000)中任意拿掉两个数，把剩下的99998个数顺序打乱，并且放入数组A中。要求只扫描一遍数组，把这两个数找出来。可以使用最多不超过5个局部变量，不能用数组变量，并且不能改变原数组的值。

first problem is discussed in [全民编程]76道高难度C++练习题.含NOI竞赛题.欢迎挑战, http://bbs.bc-cn.net/viewthread.php?tid=147967, #6.

2nd problem is a little bit harder. I have a working c++ algorithm following all the requirements stated in the problem. I would like to give some hints to you so that you can do it yourself, instead of looking at my source code.

(a) Let N=100000. Define two integers

xor_a := a_0 ^ a_1^ ... a^{N-3}
xor_1N := 1^ 2 ^3 ^ ... ^N.

Then

xor_a ^ xor_1N = m1 ^ m2,

where m1 and m2 are the two missing numbers.

(b) We need extra info to extract m1 and m2 from m1 ^ m2. You can verify that

if i ^ j == m1 ^ m2 and i * j == m1 * m2, then i == m1, j == m2 or i == m2, j == m1.

However, to get the product m1 * m2, you need some careful handling.

//***********************************
//螺旋矩阵
//***********************************
#include <iostream.h>
#include <stdio.h>
#define MAX 11

int array[MAX][MAX];
int Row;
int temp;
void GoRight(int &a,int &m)
{
    for(temp=1;temp<=Row;temp++)
        if(array[a][temp]==0)
            array[a][temp]=m++;
    a++;
}
void GoDown(int &b,int &m)
{
    for(temp=1;temp<=Row;temp++)
        if(array[temp][b]==0)
            array[temp][b]=m++;
    b--;
}
void GoLeft(int &c,int &m)
{
    for(temp=Row;temp>0;temp--)
        if(array[c][temp]==0)
            array[c][temp]=m++;
    c--;
}
void GoUp(int &d,int &m)
{
    for(temp=Row;temp>0;temp--)
        if(array[temp][d]==0)
            array[temp][d]=m++;
    d++;
}
int main()
{
    int a,b,c,d,max,m;
    cout<<\"输入维数：\";
    cin>>Row;
    for(a=1;a<=Row;a++)
        for(b=1;b<=Row;b++)
            array[a][b]=0;
    m=1,a=1,b=Row,c=Row,d=1;
    max=Row*Row;
    while(m<=max)
    {
        GoRight(a,m);
        GoDown(b,m);
        GoLeft(c,m);
        GoUp(d,m);
    }
    for(a=1;a<=Row;a++)
    {
        for(b=1;b<=Row;b++)
            printf(\"%3d\",array[a][b]);
        cout<<endl;
    }
    return 0;
}