using namespace std;
int main()
{
char str[50] = {0};
char subStr[10] = {0};
//输入M?l尾
cout<<"input the string:";
for(int i = 0;i<sizeof(str)/sizeof(str[0]);i++)
{
cin>>str[i];
if(str[i] == '#')
{
str[i] = '\0';
break;
}
}
//输入子串?l尾
cout<<"input the substring:";
for(int i = 0;i<sizeof(subStr)/sizeof(subStr[0]);i++)
{
cin>>subStr[i];
if(subStr[i] == '#')
{
subStr[i] = '\0';
break;
}
}
//用于标记是否扑ֈ子串
bool isSub = true;
//扑֭?/span>
for(int i = 0;i<strlen(str);i++)
{
isSub = true;
for(int j = 0;j<strlen(subStr);j++)
{
if(subStr[j] == str[i+j]) continue;
else
{
isSub = false;
break;
}
}
if(isSub)
{//如果扑ֈ子串Q则后l字W全部前U?/span>
for(int m = i;m<strlen(str)+1-strlen(subStr);m++)
{
str[m] = str[m+strlen(subStr)];
}
}
}
//输出删除后的
cout<<"the result string:"<<str<<endl;
system("pause");
return 1;
}
]]>
Subject: Bubble sort
Author: shexinwei
Date: 2010-09-28
Laguage: C++
IDE | Tool: GCC(DEV 5.0)
Version: 1.0
Modify Time: 2010-09-28
*/
#include <cstdlib>
#include <iostream>
#include <fstream>
using namespace std;
//排序函数
int Bublle_sort(int data[],int n);
//打印数组函数
int print(int data[],int n);
int main(int argc, char *argv[])
{
int data[] = { 2,1,5,3,5,2,6,2};
int length = sizeof(data)/sizeof(data[0]);
cout<<"初始序列:"<<endl;
print(data,length);
Bublle_sort(data,length);
cout<<"最l结?"<<endl;
print(data,length);
cout<<endl;
system("pause");
return 1;
}
int Bublle_sort(int data[],int n)
{
int tmp = 0;
for(int i = 0;i < n;i++)
{
for(int j = 0;j < n-i-1;j++)
{
if(data[j]<data[j+1])
{
tmp = data[j];
data[j] = data[j+1];
data[j+1] = tmp;
}
else continue;
}
cout<<i+1<<"ơ排序结?"<<endl;
print(data,n);
cout<<endl;
}
return 1;
}
int print(int data[],int n)
{
for(int i = 0;i < n;i++)
{
cout<<data[i]<<" ";
}
cout<<endl;
}
调试l果Q?br>
]]>
/**//////////////////////////////////////////////////2
//q雷W试题：3
//有N个大不{的自然敎ͼ1--NQ，请将它们由小到大排序?nbsp; 4
//要求E序法Q时间复杂度为O(n)Q空间复杂度为O(1)?/span>5
6
#define TEST_XUNLEI7
8
#ifdef TEST_XUNLEI9
10
#include <iostream>11
using namespace std;12
int sort(int data[],int n);13
int main()14
{15
int data[] = {8,7,9,4,6,5,3,2,1};16
for(int i = 0 ;i < sizeof(data)/sizeof(int);i++)17
cout<<data[i]<<" ";18
cout<<endl;19
sort(data,sizeof(data)/sizeof(int));20
system("pause");21
return 1;22
}23
int sort(int data[],int n)24
{25
//保证I间复杂度ؓO(1)26
int tmp = 0; 27
for(int i = 0;i < n;i++) 28
{29
//UdQ直到第data[i]为i+1的时,whilel束循环。向后l判?/span>30
while(data[i] != i+1) 31
{ 32
//每次循环保证了data[i-1]的正。d有n个所以时间复杂度为O(N)33
tmp = data[data[i]-1];34
data[data[i]-1] = data[i]; 35
data[i] = tmp;36
}37
}38
for(int i = 0;i < n;i++)39
cout<<data[i]<<" ";40
cout<<endl;41
return 1;42
}43
44
45
#endif
该题的重点在于，N?--N的数?
]]>
1/**//*
2Subject: the value of the each bit on int
3Author: shexinwei
4School: xidian university
5Date: 2010-09-13
6Laguage: C++
7IDE: visual studio 6.o
8Version: 1.0
9Modify Time: 2010-09-13
10*/
11#include <iostream>
12using namespace std;
13int main()
14{
15 int i = 0;
16 cout<<"please input the number:";
17 cin>>i;
18 char *result = new char[sizeof(int)*8];
19 int j = 1;
20 for (int k = 0;k<sizeof(int)*8;(j=j<<1),k++)
21 {
22 if ( (i&j) == 0 )
23 {
24 result[k] = '0';
25 }
26 else result[k] = '1';
27 }
28 for (int m = sizeof(int)*8-1;m >=0 ; m--)
29{
30 cout<<result[m];
31 }
32 cout<<endl;
33 delete []result;
34 return 1;
35}
/**//*2
Subject: the value of the each bit on int3
Author: shexinwei4
School: xidian university5
Date: 2010-09-136
Laguage: C++7
IDE: visual studio 6.o8
Version: 1.09
Modify Time: 2010-09-1310
*/11
#include <iostream>12
using namespace std;13
int main()14
{15
int i = 0;16
cout<<"please input the number:";17
cin>>i;18
char *result = new char[sizeof(int)*8];19
int j = 1;20
for (int k = 0;k<sizeof(int)*8;(j=j<<1),k++)21
{22
if ( (i&j) == 0 )23
{24
result[k] = '0';25
}26
else result[k] = '1';27
}28
for (int m = sizeof(int)*8-1;m >=0 ; m--)29
{30
cout<<result[m];31
}32
cout<<endl;33
delete []result;34
return 1;35
}]]>
3
4
5 /*
6
7 Subject: Quick sort
8
9 Author: shexinwei
10
11 School: xidian university
12
13 Date: 2010-09-12
14
15 Laguage: C++
16
17 IDE | Tool: GCC
18
19 Version: 1.0
20
21 Modify Time: 2010-09-12
22
23 */
24
25 #include <iostream>
26
27 using namespace std;
28
29 int sort(int begin,int end,int data[],int n);
30
31 int recurs(int begin,int end,int data[],int n);
32
33 int print(int data[],int n);
34
35 int main()
36
37 {
38
39 int data[] = {20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1};
40
41 //打印初始序列
42
43 print(data,20);
44
45 //递归排序
46
47 recurs(0,19,data,20);
48
49 //打印最l序?nbsp;
50
51 print(data,20);
52
53 cout<<endl;
54
55 return 1;
56
57 }
58
59
60
61 //打印序列
62
63 int print(int data[],int n)
64
65 {
66
67 for(int i = 0;i<n;i++)
68
69 cout<<data[i]<<" ";
70
71 cout<<endl;
72
73 return 1;
74
75 }
76
77
78
79
80
81 int sort(int begin,int end,int data[],int n)
82
83 {
84
85 //选择最后一位做初始的中?/span>
86
87 int pos = end;
88
89 int key = data[end];
90
91 //索引初始?/span>
92
93 int front = begin;
94
95 int last = end;
96
97 for(;front != last;front++)
98
99 { //向后查找
100
101
102
103 if((data[front] < key) || (data[front] == key)) continue;
104
105 else
106
107 { //遇到比中枢值大的元素，交换位置Q修改pos的|q开始从后向前找
108
109 data[pos] = data[front];
110
111 pos = front;
112
113 //以下一句可注释Q只需要保存中枢的位置Q可以不修改|因中枢值在key中保存，最后找到最l位|后再赋值即?/span>
114
115 data[pos] = key;
116
117 //输出改变一ơ后的序?/span>
118
119 print(data,n);
120
121 for(;front != last;last--)
122
123 {//从后向前查找
124
125 if((data[last] > key)||(data[last] == key)) continue;
126
127 else
128
129 { //遇到比中枢值小的元素，交换位置Q修改pos|q新开始从前向后查?/span>
130
131 data[pos] = data[last];
132
133 pos = last;
134
135 //以下一句可注释Q只需要保存中枢的位置Q可以不修改|因中枢值在key中保存，最后找到最l位|后再赋值即?/span>
136
137 data[pos] = key;
138
139 //打印修改以后的序?/span>
140
141 print(data,n);
142
143 //重新开始从前向后查?/span>
144
145 break;
146
147 }
148
149 }
150
151
152
153 }
154
155 //如果从后向前已经使得front==lastQ那么一分割完?/span>
156
157 if(front == last) break;
158
159 }
160
161 data[pos] = key;
162
163 return pos;
164
165 }
166
167 int recurs(int begin,int end,int data[],int n)
168
169 {
170
171 //递归出口
172
173 if((begin == end) || (begin > end)) return 1;
174
175 else
176
177 {
178
179 //做分?/span>
180
181 int index = sort(begin,end,data,n);
182
183 //前一部分递归
184
185 recurs(begin,index-1,data,n);
186
187 //后一部分递归
188
189 recurs(index+1,end,data,n);
190
191 }
192
193 }
]]>
]]>
作ؓ一个专业的E序员，我们需要在每一个源文g中做一些版权声明性的注释。现在根据我所见到的，和自q理解Q定义一份属于我自己的版权声明文格式。不之处日后l添加补充?/p>
CZ
|
]]>