//第一题解答
//从算式中可看出 G = 0 F = 5
//则其余 8 个属于集合 {1，2，3，4，6，7，8，9}
//生成该集合的所有排列再进行判断，利用 bExit 标志在获取结果后及时终止
//排列生成使用递归
#include <iostream>
#include <algorithm>
using namespace std;
enum Table {A, B, C, D, E, X, Y, Z};
void Compute(int List[], int First, int Last)
{
static bool bExit;
if(bExit)
return;
if(First == Last)
{
int Tmp = 1 + List[C] + 2 * List[D];
if(Tmp % 10 != List[Z])
return;
Tmp = Tmp/10;
if((List[B] + Tmp)%10 != List[Y])
return;
if(List[X] != List[A] + (List[B] + Tmp)/10)
return;
bExit = true;
cout<<List[A]<<List[B]<<List[C]<<List[D]<<List[E]<<endl
<<" "<<List[D]<<5<<0<<endl
<<"+ "<<List[D]<<5<<0<<endl
<<"-----"<<endl
<<List[X]<<List[Y]<<List[Z]<<List[D]<<List[E]<<endl;
}
else
for(int i = First; i <= Last; ++i)
{
std::swap(List[First], List[i]);
Compute(List, First + 1, Last);
if(bExit)
return;
std::swap(List[First], List[i]);
}
}
int main()
{
int List[8] = {1, 2, 3, 4, 6, 7, 8, 9};
Compute(List, 0, 7);
return 0;
}
//第二题解答
//思路和eastcowboy一样，别说我抄袭
#include<iostream>
#include<iomanip>
using namespace std;
enum {A, B, C, D, E};
bool Test(int Src, int Bit)
{
return (Src & (1<<Bit)) != 0;
}
int main()
{
char Table[] = "ABCDE";
for(int x = 1; x <= 32; ++x)
{
if(Test(x, A) && !Test(x, B))
continue;
if(Test(x, B) && Test(x, C))
continue;
if(Test(x, C) != Test(x, D))
continue;
if(!Test(x, D) && !Test(x, E))
continue;
if(Test(x, E) && !(Test(x, A) && Test(x, D)))
continue;
for(int y = 0; y < 5; ++y)
{
if(Test(x, y))
cout<<Table[y];
}
cout<<endl;
}
return 0;
}
#include <iostream>
using namespace std;
//第三题解答
//总觉得这题用递归比较好解决，于是...
//递归绘图
void _Draw(int First, int Last, const char* Fill, int Size)
{
static char* DataPtr;
char* Tmp = new char[Size+1];
if(DataPtr)
memcpy(Tmp, DataPtr, Size+1);
else
Tmp[Size] = 0;
DataPtr = Tmp;
if(First > Last)
return;
for(int i = First; i <= Last; ++i)
Tmp[i] = Fill[First];
puts(Tmp);
if(First == Last)
return;
_Draw(First+1, Last-1, Fill, Size);
puts(Tmp);
delete [] Tmp;
}
inline
void Draw(const char* Fill, int Size)
{
return _Draw(0, Size-1, Fill, Size);
}
int main()
{
const char Fill[] = "TJ12345678";
int n;
if(cin>>n && n > 3 && n < 20)
Draw(Fill, n);
return 0;
}
//第四题解答
//还不能理解什么是拉丁方阵
//位置暂留
//第五题解答
//只支持正整数的数制转换
#include <iostream>
#include <cassert>
using namespace std;
char* DecToX(char* Dst, int Src, int Base)
{
assert(Dst);
assert(Base > 0);
char Table[] = "0123456789ABCDEF";
char Buffer[sizeof(int)*8 + 1] = {0};
char* Ptr = Buffer;
char* Dest = Dst;
while(Src)
{
*Ptr++ = Table[Src%Base];
Src/=Base;
}
while(--Ptr >= Buffer)
*Dest++ = *Ptr;
*Dest = 0;
return Dst;
}
int main()
{
char Buffer[sizeof(int)*8+1];
int Num;
int Base;
if(cin>>Num && cin>>Base && Num > 0 && Base > 0 && Base <= 16)
cout<<DecToX(Buffer, Num, Base);
return 0;
}
//第六题解答
//位置暂留
//第七题解答
#include <iostream>
#include <algorithm>
#include <functional>
#include <string>
using namespace std;
void Reverse(char* First, char* Last)
{
--Last;
while(First < Last)
swap(*First++, *Last--);
}
const char Delim = '%';
int main()
{
string Tmp;
char* Buffer;
if(getline(cin, Tmp, Delim))
{
Buffer = new char[Tmp.size() + 1];
char* Last = Buffer + Tmp.size();
*Last = 0;
Tmp.copy(Buffer, Tmp.size());
char* SubFirst = Buffer;
char* SubLast;
do
{
SubFirst = find_if(SubFirst, Last, isalpha);
SubLast = find_if(SubFirst, Last, not1(ptr_fun(isalpha)));
if(SubFirst != SubLast && (*SubFirst == 'A' || *(SubLast-1) == 'N'))
Reverse(SubFirst, SubLast);
}while((SubFirst = SubLast) != Last);
cout<<Buffer<<endl;
delete [] Buffer;
}
return 0;
}
您所在位置：
GCC
[专家分：14380] 发布于 2006-04-14 11:53:00 
