笔试题练习(九)
1,有一个数组a[1000]存放0--1000;要求每隔二个数删掉一个数,到末尾时循环至开头继续进行,求最后一个被删掉的数的原始下标位置。
using namespace std;
struct Node
{
int value;
bool isDeleted;
};
int main()
{
Node a[1001];
int i,n = 12;
for (i = 0; i < n; ++i)
{
a[i].value = i;
a[i].isDeleted = false;
}
int cur = 0,count,num = n;
while (num != 1)
{
count = 0;
while (count != 2)
{
cur = (cur+1)%n;
if (a[cur].isDeleted == false)
{
count++;
}
}
a[cur].isDeleted = true;
while (a[cur].isDeleted != false)
{
cur = (cur+1)%n;
}
--num;
}
cout << cur << endl;
return 0;
}
2,有一个整数数组,现要求实现这个整数数组的循环右移。如:1,2,3,4,5 则循环右移两位后结果是:4,5,1,2,3。
using namespace std;
void shiftRight(int a[], int n,int m)
{//循环右移m位
int pre = 0,cur = (pre+m)%n,curNum = a[pre],tmp,count = 0;
if (n%2 == 0)
{//偶数
//移动偶数位
while (count != n/2)
{
tmp = a[cur];
a[cur] = curNum;
curNum = tmp;
pre = cur;
cur = (cur+m)%n;
++count;
}
//移动奇数位
count = 0;
pre = ++pre;
cur = (pre+m)%n;
curNum = a[pre];
while (count != n/2)
{
tmp = a[cur];
a[cur] = curNum;
curNum = tmp;
pre = cur;
cur = (cur+m)%n;
++count;
}
}
else
{//奇数
while (count != n)
{
tmp = a[cur];
a[cur] = curNum;
curNum = tmp;
pre = cur;
cur = (cur+m)%n;
++count;
}
}
}
int main()
{
int a[] = {1,2,3,4,5,6,7,8};
shiftRight(a,8,2);
return 0;
}
3, 以单词为最小单位翻转字符串
#include <stack>
#include <string>
using namespace std;
string reverse_string_word_by_word(string input)
{
stack<string> s;
char chSplit = ' ';
size_t pos = input.find_first_of(chSplit);
size_t lastPos = input.find_last_of(chSplit);
size_t nBegin = 0;
size_t len = 0;
string tmpStr;
size_t nWhiteSpace = 0;
while (true)
{
++nWhiteSpace;
if(pos == input.npos)
{
len = input.length() - nBegin;
tmpStr = input.substr(nBegin,len);
s.push(tmpStr);
break;
}
len = pos-nBegin;
tmpStr = input.substr(nBegin,len);
s.push(tmpStr);
nBegin = pos+1;
pos = input.find(chSplit,nBegin);
}
string result = "";
while (!s.empty())
{
string tmp = s.top();
result.append(tmp.c_str());
if (--nWhiteSpace)
{
result.append(" ");
}
s.pop();
}
return result;
}
int main()
{
string strData = "the house is blue";
string result = reverse_string_word_by_word(strData);
cout << result << endl;
return 0;
}
4,题目描述:
设有n个正整数,将它们联接成一排,组成一个最大的多位整数。
例如:
n=2时,2个整数32,321连接成的最大整数为:32321,
n=4时,4个整数55,31,312, 33 联接成的最小整数为:553332131312
#include <vector>
#include <algorithm>
using namespace std;
class MyCompator
{
public:
bool operator()(char* lhsStr,char* rhsStr)
{
if (lhsStr == NULL)
return false;
if (rhsStr == NULL)
return true;
const char* temps1 = lhsStr;
const char* temps2 = rhsStr;
while(!((*temps1 == 0) && (*temps2 == 0)))
{
// 字符串相加的效果
if(*temps1 == 0)
{
temps1 = rhsStr;
}
if(*temps2 == 0)
{
temps2 = lhsStr;
}
// 比较
if(*temps1 != *temps2)
{
return *temps1 > *temps2;
}
else
{
temps1++;
temps2++;
}
}
// 两个相等
return true;
}
};
int main()
{
char* ps[] = { "55","31","312","33","321"};
vector<char*> v(ps,ps+sizeof(ps)/sizeof(*ps));
sort(v.begin(), v.end(), MyCompator());
copy(v.begin(), v.end(), ostream_iterator<char*>(cout, ""));
cout << endl;
return 0;
}
5、编程题
输入:N(整数)
输入:数据文件A.txt,不超过6条记录,字符串长度不超过15个字节
文件格式如下:
字符串"t数字"n
说明:
每行为1条记录;字符串中不含有"t。
数字描述的是该字符串的出现概率,小于等于100的整数。
多条记录的出现概率之和为100,如果A.txt不满足该条件,程序则退出;
如果文件格式错误,程序也退出。
要求:
编写一个程序,输入为N(正整数),读入文件A.txt,按照字符串出现概率随机地输出字符串,输出N条记录
例如:
输入文件A.txt
abc"t20
a"t30
de"t50
输入为:10
即 abc有20%的概率输出,a有30%的概率输出,de有50%的概率输出,输出10条记录
以下为一次输出的结果,多次输出的结果可能不相同。
abc
a
de
de
abc
de
a
de
a
de
import java.io.FileInputStream;
import java.io.InputStreamReader;
import java.util.Random;
import java.util.Scanner;
import java.util.Vector;
//记录类
class Record
{
private String name = null;
private int num = 0;
public Record(String name, int num)
{
this.name = name;
this.num = num;
}
public String getName()
{
return name;
}
public void setName(String name)
{
this.name = name;
}
public int getNum()
{
return num;
}
public void setNum(int num)
{
this.num = num;
}
}
public class test
{
public static void main(String[] args) throws Exception
{
int N,i;
Scanner scanner = new java.util.Scanner(System.in);
N = scanner.nextInt();
FileInputStream input = null;
InputStreamReader bufferInput = null;
BufferedReader reader = null;
Vector<Record> records = new java.util.Vector<Record>();
try
{
input = new FileInputStream("D:\\data1.txt");
bufferInput = new InputStreamReader(input);
reader = new BufferedReader(bufferInput);
String line = "";
String strError = "\\t";
String strSpliter = "\t";//分隔符
int total = 0;
while ((line = reader.readLine()) != null)
{
if (line.indexOf(strError) != -1)
{//含有\t
System.exit(1);
}
String[] record = line.split(strSpliter);
if (record.length != 2)
{//记录格式错误
System.exit(1);
}
String name = "";
int num = 0;
try
{
name = record[0];
num = Integer.parseInt(record[1]);
if (num > 100)
{//单个记录超过100
System.exit(1);
}
total += num;
}
catch(java.lang.Exception ex)
{//解析错误
System.exit(1);
}
Record node = new Record(name,num*N/100);
records.add(node);
}
if (total != 100)
{//概率和不为100
System.exit(1);
}
int size = records.size();
int pos = -1,count = 0;
for (i = 0; count != N; ++i)
{
Random random = new java.util.Random();
//在记录集合中随机取一条记录来检验
pos = random.nextInt(size);
Record temp = records.elementAt(pos);
if (temp.getNum() != 0)
{//若此记录可以被输出(次数还没减到0)
//输出记录并将其次数减去1
System.out.println(temp.getName());
count++;
temp.setNum(temp.getNum()-1);
}
}
}
catch(java.lang.Exception ex)
{
System.exit(1);
}
finally
{//关闭文件
input.close();
bufferInput.close();
reader.close();
}
}
}
作者:洞庭散人
出处:http://phinecos.cnblogs.com/
posted on 2009-06-04 14:29 Phinecos(洞庭散人) 阅读(1658) 评论(1) 编辑 收藏 举报