（测试环境为windows8下vs2013）
3-0编译运行测试本章程序
#include <iomanip>
#include <ios>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
cout << "please enter your name: ";
string name;
cin >> name;
cout << "hello," << name << endl;
cout << "please enter your midterm and final exam grades: ";
double midterm ,final;
cin >> midterm >> final;
cout << "enter all your homework grades,"
"followed by end-of-file";
vector<double> homework;
double x;
while (cin >> x){
homework.push_back(x);
}
typedef vector<double>::size_type vec_sz;
vec_sz size = homework.size();
if (size == 0){
cout << endl << "no grade,please try again!" << endl;
return 1;
}
sort(homework.begin(), homework.end());
vec_sz mid = size / 2;
double median;
median = size % 2 == 0 ? (homework[mid] + homework[mid + 1]) / 2 : homework[mid];
streamsize prec = cout.precision();
cout << "you final grade is " << setprecision(3)
<< 0.2*midterm + 0.4*final + 0.4*median
<<setprecision(prec)<< endl;
return 0;
}
文件结束：Ctrl+Enter+z
prec:重新恢复cout的精度
3-2写一个程序计算并输出一个整数集的四位分数（也就是，最大值的1/4，然后是下一个最大值的1/4，如此而已）
（四分位数：即统计学中，把所有数值由小到大排列并分成四等份，处于三个分割点位置的数值就是四分位数。）
相关算法如下：
1.将n个数从小到大排列：
2.Q2为n个数组成的数列的中数；
3.当n为奇数时，中数Q2将该数列分为数量相等的两组数，每组有 (n-1)/2 个数，Q1为第一组 (n-1)/2 个数的中数，Q3为为第二组(n+1)/2个数的中数；
当n为偶数时，中数Q2将该数列分为数量相等的两组数，每组有n/2数，Q1为第一组 n/2个数的中数，Q3为为第二组 n/2 个数的中数。
#include <iomanip>
#include <ios>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
cout << "please input the set of integer :" << endl;
vector<double> sets;
double x;
while (cin >> x){
sets.push_back(x);
}
typedef vector<double>::size_type vec_sz;
vec_sz size = sets.size();
vec_sz mid = size / 2;
if (size == 0){
cout << "no number! " << endl;
return 1;
}
sort(sets.begin(), sets.end());
double q1, q2, q3;
if (size % 2 == 0){
q2 = (sets[mid] + sets[mid - 1]) / 2;
}else{
q2 = sets[mid];
}
if (mid % 2 == 0){
q1 = (sets[mid / 2] + sets[mid / 2 - 1]) / 2;
q3 = (sets[size - 1 - mid / 2] + sets[size - mid / 2]) / 2;
}else{
q1 = sets[mid / 2];
q3 = sets[size - 1 - mid / 2];
}
streamsize prec = cout.precision();
cout << "the first block is: " <<setprecision(3)<< q1 << endl;
cout << "the second block is: " << q2 << endl;
cout << "the third block is: " << q3 <<setprecision(prec)<< endl;
return 0;
}
测试结果：
3-3写一个程序，计算输入中每个不同的单词出现了多少次
算法思路：
1.用一个vector<string>存储输入文本中所有的单词
2.用另一个vector<string>存储不同的单词，即重复的就没有了
3.用一个int变量输出单词出现的次数（要存储的话用vector即可）
#include <iomanip>
#include <ios>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
cout << "please input the text: " << endl;
vector<string> text;
string word;
while (cin >> word){
text.push_back(word);
}
if (text.size() == 0){
cout << "no text! please try again: " << endl;
return 1;
}
// put the different word to vector for temp
vector<string> temp;
int i, j;
for (i = 0; i != text.size(); i++){
for (j = 0;; j++){
if (j == temp.size()){
temp.push_back(text[i]);
break;
}else{
if (temp[j] == text[i])
break;
}
}
}
//output the count of word
int count;
for (i = 0; i != temp.size(); i++){
count = 0;
for (j = 0; j != text.size(); j++){
if (temp[i] == text[j]){
count++;
}
}
cout << "the word: " << temp[i] << " present " << count << " times" << endl;
}
return 0;
}
运行结果：
3-4写一个程序，找出输入中最长和最短的字符串
#include <iomanip>
#include <ios>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
cout << "please input the text: " << endl;
vector<string> text;
string word;
while (cin >> word){
text.push_back(word);
}
if (text.size() == 0){
cout << "no text! please try again: " << endl;
return 1;
}
//the strmax use for store the longest string
//the max use for store the length of the longest string
//the strmin and the min also
string strmax, strmin;
string::size_type max, min;
strmax = strmin = text[0];
max = min = text[0].size();
for (int i = 1; i != text.size(); i++){
if (text[i].size() > max){
max = text[i].size();
strmax = text[i];
}else if (text[i].size() < min){
min = text[i].size();
strmin = text[i];
}else
continue;
}
cout << "the longest word is: " << strmax <<endl
<< " and the length is: " << max << endl;
cout << "the shortest word is " << strmin <<endl
<<" and the length is: " << min << endl;
return 0;
}
测试结果：
3-5写一个程序，可以一次计算多个学生的成绩，这个程序需要同时使用两个vector：
第一个vector要保存学生的姓名，第二个vector要保存根据输入计算出的最终成绩。
假定家庭作业成绩的数目是确定的（本文假定为3，计算的为平均成绩）。
#include <iomanip>
#include <ios>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
vector<string> name;
vector<double> grade;
string sname;
double x;
while (cin >> sname){
name.push_back(sname);
double sum = 0;
int count = 0;
while (cin >> x){
sum += x;
count++;
if (count == 3)
break;
}
grade.push_back(sum / count);
}
if (name.size() == 0){
cout << "no student ,try again!" << endl;
return 1;
}
streamsize prec = cout.precision();
for (int i = 0; i != name.size(); i++){
cout << "the student: " << name[i]
<< setprecision(3) << " ‘s grade is :"
<< grade[i] << setprecision(prec)
<< endl;
}
return 0;
}
运行结果：
3-6在3.1节程序中，如果学生没有输入任何家庭作业成绩，那么计算平均值时
可能会除以0，在c++中，除以0是不确定的，也就是说系统会进行随意的操作
在你的c++系统中会怎么样呢？重写这个程序，使得程序的行为不依赖于系统
如何对待除以0的情况。
出现情况：
在程序中加入检查代码即可，如本文3-0程序
if (size == 0){
cout << endl << "no grade,please try again!" << endl;
return 1;
}
再次出现时：
时间： 2024-11-07 07:12:16