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金山训练营入学考试题
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1 编写函数实现十进制正整数到十四进制数的转换，在屏幕输出转换结果。
说 明：用0, 1, 2, 3,....., 8, 9, A, B, C, D表示十四进制的基本的14个数。
例：键盘输入14，屏幕输出10。
2 结构RECT可以表示一个平面上的矩形区域 ：
struct RECT
{
int left, top, right, bottom;
};
如果给定两个矩形区域A和B，请构造一个函数，用若干矩形区域表示出区域A被B切掉之后剩余的区域，并在屏幕输出结果。例如，区域 (0, 0)-(2, 2) 被 (1, 1)-(3, 3) 切掉之后，剩余的两个区域可以表示为 (0, 0)-(2, 1) 和 (0, 1)-(1, 2)，或者是 (0, 0)-(1, 2) 和 (1, 0)-(2, 1)（两种表示方式均可）。
3 文件名：3.cpp
功 能：编程实现猜词游戏
说 明：对于单词“hello”，程序提示输出：?????，等待用户输入。
用户输入时，若单词包含该字母，如“l”，则程序显示输出“??ll?”；
若单词不含该字母，如“a”，则程序提示用户猜错。
继续等待用户输入，直到用户猜出全部字母，或输入错误次数超过最大允许出错次数，游戏结束。
条 件：1) 单词由程序内定，由全小写字母组成
2) 提示输出问号数量等于单词长度
3) 最大允许出错次数等于单词长度
哪位高手写一下代码....小弟做了..感觉不是很好...想看一下高手是怎么写的??
========================================================================
1:
void fun( int input)
{
if (input >= 14)
fun(input/14);
printf("%c","0123456789ABCD"[input%14]);
}
2.0
#include <iostream>
using namespace std ;
class Rect
{
public:
Rect(int x, int y , int m ,int n)
{
left = x ;
top = y ;
right = m ;
bottom = n ;
}
int left, top, right, bottom;
};
void printArea(int x, int y , int m ,int n)
{
cout < <"(" < <x < <"," < <y < <")" < <"-" < <"(" < <m < <"," < <n < <")" < <endl;
}
void intersection(const Rect &r1, const Rect &r2)
{
//有15种相切情况,穷举每一种情况。
if(r2.bottom < r1.bottom && r2.bottom > r1.top)
{
if(r2.left <= r1.left && r2.right >= r1.right && r2.top <= r1.top)
{
printArea(r1.left , r2.bottom, r1.right , r1.bottom) ;//1
return ;
}
else if(r2.left <= r1.left && r2.right >= r1.right && r2.top > r1.top)
{
printArea(r1.left , r1.top , r1.right, r2.top) ;//2
printArea(r1.left , r2.bottom, r1.right , r1.bottom) ;
return ;
}
else if(r2.right < r1.right && r2.right > r1.left && r2.left <= r1.left && r2.top <= r1.top)
{
printArea(r1.left , r2.bottom , r2.right , r1.bottom) ;//3
printArea(r2.right ,r1.top , r1.right ,r1.bottom ) ;
return ;
}
else if(r2.left > r1.left && r2.left < r1.right && r2.right >= r1.right && r2.top <= r1.top)
{
printArea(r1.left ,r1.top , r2.left ,r1.bottom) ;//4
printArea(r2.left ,r2.bottom ,r1.right ,r1.bottom) ;
return ;
}
else if(r2.left > r1.left && r2.right < r1.right && r2.top <= r1.top)
{
printArea(r1.left,r1.top , r2.left ,r2.bottom) ;//5
printArea(r2.right , r1.top , r1.right ,r2.bottom) ;
printArea(r1.left , r2.bottom , r1.right ,r1.bottom) ;
return ;
}
}
if(r2.top > r1.top && r2.top < r1.bottom && r2.bottom >= r1.bottom )
{
if(r2.left <= r1.left && r2.right >= r1.right )
{
printArea(r1.left ,r1.top ,r1.right ,r2.top) ;//6
return ;
}
else if(r2.right < r1.right && r2.right > r1.left && r2.left <= r1.left)
{
printArea(r1.left , r1.top ,r2.left , r1.bottom) ;//7
printArea(r2.left ,r2.bottom ,r1.right ,r1.bottom) ;
return;
}
else if(r2.left > r1.left && r2.left < r1.right && r2.right >= r1.right)
{
printArea(r1.left ,r1.top , r2.left , r1.bottom) ;//8
printArea(r2.left , r1.top ,r1.right ,r2.top) ;
return ;
}
else if(r2.left > r1.left && r2.right < r1.right )
{
printArea(r1.left ,r1.top ,r1.right ,r2.top) ;//9
printArea(r1.left ,r2.top ,r2.left , r1.bottom) ;
printArea(r2.right ,r2.top ,r1.right ,r1.bottom) ;
return ;
}
}
if(r2.right > r1.left && r2.right < r1.right)
{
if(r2.top <= r1.top && r2.bottom >= r1.bottom && r2.left <= r1.left)
{
printArea(r2.right ,r1.top ,r1.right,r1.bottom) ;//10
return ;
}
else if(r2.top <= r1.top && r2.bottom >= r1.bottom && r2.left > r1.left)
{
printArea(r1.left ,r1.top ,r2.left ,r1.bottom) ;//11
printArea(r2.right ,r1.top ,r1.right ,r1.bottom) ;
return ;
}
else if(r2.top > r1.top && r2.bottom < r1.bottom && r2.left <= r1.left)
{
printArea(r1.left ,r1.top ,r2.right ,r2.top) ;//12
printArea(r1.left ,r2.bottom ,r2.right , r1.bottom) ;
printArea(r2.right , r1.top ,r1.right ,r1.bottom) ;
return ;
}
}
if(r2.left < r1.right && r2.left >r1.left && r2.right >= r1.right)
{
if(r2.top <= r1.top && r2.bottom >= r1.bottom )
{
printArea(r1.left ,r1.top , r2.left , r2.bottom) ;//13
return ;
}
else if(r2.top > r1.top && r2.bottom < r1.bottom)
{
printArea(r1.left , r1.top ,r2.left , r1.bottom) ;//14
printArea(r2.left ,r1.top ,r1.right ,r2.top) ;
printArea(r2.left ,r2.bottom , r1.right ,r1.bottom) ;
return ;
}
}
if(r2.left > r1.left && r2.right < r1.right && r2.top > r1.top && r2.bottom < r1.bottom)
{
printArea(r1.left ,r1.top ,r2.left ,r2.top) ;//15
printArea(r2.left,r1.top ,r2.right ,r2.top) ;
printArea(r2.right,r1.top ,r1.right ,r2.top) ;
printArea(r1.left , r2.top ,r2.left ,r2.bottom) ;
printArea(r2.right ,r2.top ,r1.right ,r2.bottom) ;
printArea(r1.left , r2.bottom , r2.left , r1.bottom) ;
printArea(r2.left ,r2.bottom , r2.right , r1.bottom) ;
printArea(r2.right , r2.bottom , r1.right,r1.bottom) ;
return ;
}
cout < <"不相切！" < <endl;
}
int main()
{
Rect r1(0,0,2,2) ;
Rect r2(1,1,3,3) ;
//Rect r1(0,0,3,3) ;
//Rect r2(1,1,2,2) ;
//Rect r1(0,0,3,3) ;
//Rect r2(0,1,3,2) ;
intersection(r1,r2) ;
return 0 ;
}
3.0
#include <iostream>
using namespace std;
#include <string>
void guessword(char * srcword){
int len = strlen(srcword);
if(len==0)
return;
char * buf = new char [len+1];
memset(buf, 0x00, len+1);
int i;
for(i=0; i<len; i++){
buf[i] = '?';
}
cout<<buf<<endl;
int errortime = 0;
do{
char chuser;
cin>>chuser;
bool errorflag = true;
for(i=0; i<len; i++){
if(srcword[i] == chuser){
errorflag = false;
buf[i] = chuser;
}
}
cout<<buf<<endl;
if(errorflag) // 说明这次猜错了
errortime ++;
if(errortime == len){
cout<<"Error! You have tried so many times!"<<endl;
break;
}
for(i=0; i<len; i++){
if(buf[i]=='?')
break;
}
if(i==len){
cout<<"Congratulations! You are right!"<<endl;
break;
}
}while(1);
delete [] buf;
}
void main(){
char * srcword = "appreciate";
guessword(srcword);
}
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原文地址
http://topic.csdn.net/u/20080517/21/8606a5d6-9c07-4ebc-a7bb-243af402e20b.html?seed=1938171927
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发表于： 2008-05-19，修改于： 2008-05-19 15:30 已浏览492次，有评论1条
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本站网友 | 时间：2008-06-16 14:46:05 IP地址：116.60.129.★ |
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呵呵，看了看题目，有点意思，感觉你第3题做的还行，我就把前两题做了一下，代码水平有限，别介意，只是觉得思路和你有很大不同，有什么问题可以加我QQ，共同讨论：QQ：15483930
用的vc6.0编译成功
第1题：
#include <string>
#include <iostream>
#include <stdlib.h>
using namespace std;
void show_tento14(int n)
{
int r;//余数
string out14num;//要输出的14进制数的倒置
char temps[2];
while(n>0)//辗转相除法
{
r = n%14;
n /= 14;
switch(r)
{
case 10:
out14num += 'A';
break;
case 11:
out14num += 'B';
break;
case 12:
out14num += 'C';
break;
case 13:
out14num += 'D';
break;
default:
out14num += itoa(r, temps, 10);
break;
}
}
//输出
for (int i=out14num.size()-1; i>=0; --i)
{
cout<<out14num[i];
}
cout<<endl;
}
int main()
{
int n = 14;
show_tento14(n);
return 0;
}
第2题：
#include <iostream>
using namespace std;
struct rect
{
int left;
int top;
int right;
int down;
rect(int l = 0, int t = 0, int r = 0, int d = 0)
{
left = l;
top = t;
right = r;
down = d;
}
void setrect(int l, int t, int r,int d)
{
left = l;
top = t;
right = r;
down = d;
}
//判断是否为区域矩形
bool IsRect()
{
if (left>=right || top>=down)
{
return false;//为空
}
else
{
return true;
}
}
//将矩形设置为非区域矩形，即使其分量置零
void SetZero()
{
left = top = right = down;
}
bool IsEqualRect(rect &B)
{
if (left == B.left &&
top == B.top &&
right == B.top &&
down == B.down)
{
return true;
}
else
{
return false;
}
}
//求出矩形A和B的公共区域
rect commonrect(rect &B)
{
rect C;
C.left = (left>B.left ? left:B.left);
C.top = (top>B.top ? top:B.top);
C.right = (right<B.right ? right:B.right);
C.down = (down<B.down ? down:B.down);
return C;
}
void display()
{
if (!IsRect())
{
return;//若为非区域，直接返回
}
else
{
cout<<"("<<left<<","<<top<<") "
<<"("<<right<<","<<down<<")"<<endl;
}
}
};
//表示一个矩形减去其内部一块矩形后的四个矩形，若相应的分矩形为0矩形，表示没有这块
struct last_region
{
rect A;
rect B;
rect C;
rect D;
void display()
{
A.display();
B.display();
C.display();
D.display();
}
};
//求出一个矩形减去其内部一个矩形后剩余的矩形
last_region LastRegion(rect R, rect subR)
{
last_region lastregion;
if (!subR.IsRect())//如果sunR为空
{
lastregion.A = R;
lastregion.B.SetZero();
lastregion.C.SetZero();
lastregion.D.SetZero();
return lastregion;
}
lastregion.A.setrect(R.left, R.top, R.right, subR.top);
lastregion.B.setrect(R.left, subR.top, subR.left, subR.down);
lastregion.C.setrect(subR.right, subR.top, R.right, subR.down);
lastregion.D.setrect(R.left, subR.down, R.right, R.down);
return lastregion;
}
//显示A被B切割后的区域
void show_region(rect A, rect B)
{
rect C = A.commonrect(B);//公共部分
last_region lastregion = LastRegion(A, C);
lastregion.display();
}
int main()
{
rect r1(0,0,2,2);
rect r2(1,1,3,3);
//Rect r1(0,0,3,3) ;
//Rect r2(1,1,2,2) ;
//Rect r1(0,0,3,3) ;
//Rect r2(0,1,3,2) ;
show_region(r1, r2);
return 0;
}
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Blog作者的回复:
呵呵..这个贴是综合csdn上N个人的成果.并非我原创的.只是觉得有学习价值.就贴过来了.谢谢你的热心回复..
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