上题分析编码规则：　
“8”(“日”) 拆成“口”和“凵”
其中　“口”　按笔划依次是 1,2,3,7号位；“凵”按笔划依次为 6,5,4号位

运行结果给出：

0 7 3 1 4 9 5 6 8 2
2 8 0 7 3 1 4 9 5 6
2 8 3 7 0 1 4 9 5 6
2 8 4 1 0 7 3 9 5 6
2 8 6 5 9 3 7 0 1 4
2 8 6 5 9 4 1 0 7 3
2 8 6 5 9 4 1 3 7 0
3 7 0 1 4 9 5 6 8 2
4 1 0 7 3 9 5 6 8 2
6 5 9 3 7 0 1 4 8 2
6 5 9 4 1 0 7 3 8 2
6 5 9 4 1 3 7 0 8 2
共有 12 种方法.

Terminated with return code 0
Press any key to continue ...

[此贴子已经被作者于2007-8-23 16:06:24编辑过]

48. 将４个红球，３个白球与３个黄球排成一排，共有多少种排法？

输出结果：

共有 4200 种排法。

Terminated with return code 0
Press any key to continue ...

算不算投机

[此贴子已经被作者于2007-8-23 16:24:55编辑过]

原来写了在9页,上次在网吧写的,没编译,现在用编译器运行下,再发上来完整的.
14.(已确定正确):
#include<iostream>
using namespace std;
int main()
{
const int n=20;
int i,j;
char t,a[n];
for(i=0;i<n/2;i++)
{
a[i]='*';
}
for(i=n/2;i<n;i++)
{
a[i]='0';
}
cout<<"原排列为:"<<endl;
for(i=0;i<n;i++)
{
cout<<a[i]<<" ";
}
for(i=n/2;i<n;i++) //利用两个for，原理很简单，我不会把别人的水平看太低，所以不写了。
for(j=i;j>0&&a[j-1]!='0';j--)
{
t=a[j];
a[j]=a[j+1];
a[j+1]=t;
}
cout<<"\n\n\n交换后排列变为:"<<endl;
for(i=0;i<n;i++)
{
cout<<a[i]<<" ";
}
cout<<endl<<endl;
return 0;
}

[此贴子已经被作者于2007-8-23 20:32:54编辑过]

Thanks for pointing out my mistake.

I still remember a thing happened when I studied in Beijing. I pointed out a mistake in a woman professor's book, she got furious.

Great algorithm.

To HJin:

It's so great and lucky that you did not do that.

[此贴子已经被作者于2007-8-28 21:29:21编辑过]

题目真多，慢慢做来看看

路过，写个第二题:

#include<iostream>
using namespace std;

bool check(int a[])
{
if(a[0]==1&&a[1]==0) return false;
if(a[1]==1&&a[2]==1) return false;
if((a[2]==1&&a[3]==0) ||(a[2]==0&&a[3]==1)) return false;
if(a[3]==0&&a[4]==0) return false;
if(a[4]==1&&(a[1]==0||a[3]==0)) return false;
return true;
}
int main()
{
int a[5];
for(a[0]=0;a[0]<2;a[0]++)
{
if(a[1]=1) a[1]=0;
for(a[1]=0;a[1]<2;a[1]++)
{
if(a[2]=1) a[2]=0;
for(a[2]=0;a[2]<2;a[2]++)
{
if(a[3]=1) a[3]=0;
for(a[3]=0;a[3]<2;a[3]++)
{
if(a[4]=1) a[4]=0;
for(a[4]=0;a[4]<2;a[4]++)
{
if(check(a))
cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<endl;
}
}
}
}
}
return 0;
}

第三题:


#include<iostream>
#define max 100
using namespace std;

void copy1(int a[],int k,int n)
{
for(int i=k+1;i<=(n-4)/2;i++)
a[i]=a[k];
}
void copy2(int a[],int k,int n)
{
for(int i=k-1;i>(n-4)/2;i--)
a[i]=a[k];
}
int main()
{
int n;
cout<<"输入n的值(n>=4):"<<endl;
cin>>n;
int a[100];
int i,j,k;
for(k=0;k<n;k++) cout<<"T"<<" ";
cout<<endl;
cout<<"T"<<" ";
for(k=0;k<n-2;k++) cout<<"J"<<" ";
cout<<"T"<<endl;
for(i=1;i<=(n-4)/2+1;i++)
{
for(j=1;j<=i;j++)
{
a[j-1]=j;
copy1(a,(j-1),n);
}
for(j=1;j<=i;j++)
{
a[n-4-j]=j;
copy2(a,(n-4-j),n);
}
cout<<"T"<<" "<<"J"<<" ";
for(k=0;k<n-4;k++) cout<<a[k]<<" ";
cout<<"J"<<" "<<"T"<<" ";
cout<<endl;
}
for(i=n-4;i>(n-4)/2+1;i--)
{
for(j=1;j<=i-((n-4)/2+1);j++)
{
a[j-1]=j;
copy1(a,(j-1),n);
}
for(j=1;j<=i-((n-4)/2+1);j++)
{
a[n-4-j]=j;
copy2(a,(n-4-j),n);
}
cout<<"T"<<" "<<"J"<<" ";
for(k=0;k<n-4;k++) cout<<a[k]<<" ";
cout<<"J"<<" "<<"T"<<" ";
cout<<endl;
}
cout<<"T"<<" ";
for(k=0;k<n-2;k++) cout<<"J"<<" ";
cout<<"T"<<endl;
for(k=0;k<n;k++) cout<<"T"<<" ";
cout<<endl;
return 0;
}

#include<iostream.h>
#define N 20
void main()
{ int i,j,k,q;
q=N/2;
char a[N][N];
for(i=0;i<N;i++)
for(j=0;j<N;j++)
{
a[i][j]='T';
}

for(i=1;i<N-1;i++)
for(j=1;j<N-1;j++)
{
a[i][j]='J';
}
char t,h=2;
t=N-2;

for(k=1;k<=q;k++)
{
for(i=h;i<t;i++)
for(j=h;j<t;j++)
{
a[i][j]=k+48;
}
t--;
h++;}
for(i=0;i<N;i++)
{ for(j=0;j<N;j++)
{cout<<a[i][j];
}
cout<<"\n";}

}