第一次发贴，做一做第1题。附件下载后改名为1.cpp

解体算法最原始的方法：穷举 （循环次数绝对不是10！）
耗时大概1S左右吧， 没有做时间测试。

虽然，B F G 可以推算出 为 9， 5 ，0 ，但还是最原始的方法让他们参加循环。
{
ABCDE
DFG
DFG
---------
XYZDE
推算B， 因为 X = A + 1 , 则 B 必须为 9。
(2 * FG ) % 100 必须为0 且F != G ，则 F = 5 G = 0;
}

下面给出算法：

//1.cpp
// hkice August 17 2007
// I use bcc5.5 free command line compiler
// c:> bcc32 1.cpp
// c:> 1.exe
//Result:abcdefgxyz
// 2978650314
//=======================
// 29786
// 850
// + 850
//-----------------------
// 31486

#include <iostream.h>

int main()
{
for (int z = 1; z < 10; z++)
{
for (int y = 0; y < 10; y++)
{
if (y == z)
continue;
for (int x = 1; x < 10; x++)
{
if (x == y || x == z)
continue;
for (int g = 0; g < 10; g++)
{
if (g == x || g == y || g == z)
continue;
for (int f = 0; f < 10; f++)
{
if (f == g || f == x || f == y || f == z)
continue;
for (int e = 0; e < 10; e++)
{
if (e == f || e == g || e == x || e == y || e == z)
continue;
for (int d = 1; d < 10; d++)
{
if (d == e || d == f || d == g || d == x || d == y || d == z)
continue;
for (int c = 0; c < 10; c++)
{
if (c == d || c == e || c == f || c == g || c == x || c == y || c == z)
continue;
for (int b = 0; b < 10; b++)
{
if (b == c || b == d || b == e || b == f || b == g || b == x || b == y || b == z)
continue;
for (int a = 1; a < 10; a++)
{
if (a == b || a== c || a == d || a == e || a == f || a == g || a == x || a == y || a == z)
continue;
//cout <<"Test" << a << b << c << d << e << f << g << x << y << z << endl;

int abcde = a * 10000 + b * 1000 + c * 100 + d * 10 + e;
int dfg = d * 100 + f * 10 + g;
int xyzde = x* 10000 + y * 1000 + z* 100 + d * 10 + e;
if ( abcde + 2 * dfg - xyzde == 0)
{
cout << "Result:" << 'a' << 'b' << 'c' << 'd' << 'e' << 'f' << 'g'
<< 'x' << 'y' << 'z' << endl;
cout << " " << a << b << c << d << e << f << g << x << y << z << endl;
cout << "=======================" <<endl;
cout << " " << abcde <<endl;
cout << " " << dfg <<endl;
cout << " + " << dfg <<endl;
cout << "-----------------------" << endl;
cout << " " << xyzde << endl;
}

}
}
}
}

}
}
}
}
}
}
return 0;
}

[此贴子已经被作者于2007-8-17 15:30:55编辑过]

第五题，有位同志给出了堆栈算法，这里给出的是递归算法：

//5.cpp
// hkice August 17 2007
// I use bcc5.5 free command line compiler
// c:> bcc32 5.cpp
// c:> 5.exe
//
//c:>bcc32 5.cpp
//Borland C++ 5.5.1 for Win32 Copyright (c) 1993, 2000 Borland
//5.cpp:
//Turbo Incremental Link 5.00 Copyright (c) 1997, 2000 Borland

//c:>5.exe
//Intput number:10
//Input ruler:3
//10(10) has converted to 101(3)

//c:>5.exe
//Intput number:3
//Input ruler:2
//3(10) has converted to 11(2)

//c:>5.exe
//Intput number:255
//Input ruler:2
//255(10) has converted to 11111111(2)

#include <iostream.h>

int convert(int number, int ruler);

int main()
{
int number, ruler;
cout << "Intput number:";
cin >> number;
cout << "Input ruler:";
cin >> ruler;
cout << number <<"(10) has converted to ";
convert(number, ruler);
cout << "(" << ruler << ")" << endl;

return 0;
}

int convert(int number, int ruler)
{
if (number >= ruler)
convert(number/ruler , ruler);
cout << number%ruler;
return number%ruler;
}

[此贴子已经被作者于2007-8-17 17:20:21编辑过]

不会啊``

第三题 交流一下
分析：可以看到方阵是中心对称的，先考虑左上角，发现a[i][j]所在层数即i，j中较小的那一个（层数从零算起）这样就可以为左上角赋值，后边就好解决了
#include <stdio.h>
#define N 15
main()
{
char a[N][N]={0};
int i,j,M,storey,tag=N%2;
storey=(N+1)/2-1;
for(i=0;i<=storey;i++)
for(j=0;j<=storey;j++)
{
M=(i<=j)?i:j;
if(0==M) a[i][j]='T';
else if(1==M) a[i][j]='J';
else a[i][j]=M-1+'0';
}
for(i=0;i<=storey;i++)
for(j=storey+1;j<N;j++)
{
if(tag)
a[i][j]=a[i][2*M-j];
else
a[i][j]=a[i][2*M-j+1];
}
for(i=storey+1;i<N;i++)
for(j=0;j<N;j++)
{
if(tag)
a[i][j]=a[2*M-i][j];
else
a[i][j]=a[2*M-i+1][j];
}
for(i=0;i<N;i++)
{
for(j=0;j<N;j++)
printf("%c ",a[i][j]);
printf("\n");
}
}

关于第10题62楼可能有错，个数对得上，就总面积不行
请高手指点

如图１所示，编写程序计算 ┎┰┰┰┰┰┰┰┰┰┒
大大小小正方形共有多少？当最小 ┠╂╂╂╂╂╂╂╂╂┨
正方行边长为１时，它们的总面积 ┠╂╂╂╂╂╂╂╂╂┨
共为多少？ ┠╂╂╂╂╂╂╂╂╂┨
┠╂╂╂╂╂╂╂╂╂┨
┠╂╂╂╂╂╂╂╂╂┨
┠╂╂╂╂╂╂╂╂╂┨
┠╂╂╂╂╂╂╂╂╂┨
┠╂╂╂╂╂╂╂╂╂┨
┠╂╂╂╂╂╂╂╂╂┨
┖┸┸┸┸┸┸┸┸┸┚



#include<iostream>
using namespace std;
int main()
{
int a[11][11];
int q;
int n=0;
int s=0;
cout<<"请输入第一边有多少点=";
cin>>q;
for(int i=0;i<q;i++)
for(int j=0;j<q;j++)
for(int k=0;k<q;k++)
for(int m=0;m<q;m++)
if((i-k)&&(i-k)==(m-j)){n+=1;s+=(i-k)*(i-k);}
cout<<"正方行总个数n="<<n/2.0<<endl;
cout<<"总面积s="<<s/2.0<<endl;
cin>>q;
}

[此贴子已经被作者于2007-8-21 20:41:51编辑过]

Great minds think alike!

YOU made a mistake at the red part
I corrected it behind in green.

[此贴子已经被作者于2007-8-22 16:35:10编辑过]

嘿嘿```大家好,好久不见,闭关练武,才回来```

55. (液晶显示) 下图是用液晶七笔阿拉数字表示的十个数字，我们把横和竖的一
个短划都称为一笔，即７有３笔，８有７笔等。请把这十个数字重新排列，要做到
两相邻数字都可以由另一个数字加上几笔或减去几笔组成，但不能又加又减。比如
７→３是允许的，７→２不允许。编程打印出所有可能的排列。
如：４１０７３９５６８２。