oadjing的专栏

函数模板重载

程序说明：如果 kingLear.txt 行与行之间有单个或多个空行，则忽略空行内容，继续向下读取文件内容并显示。 不论文件末尾最后一行有无换行符程序均可正常运行。 kingLear.txtIs not this your son, my lord?I cannot conceive you.I cannot wish the fault undone

fin.txt#include <iostream>#include <fstream>using namespace std;void newLine(istream& in_stream);int main() { ifstream fin; cout << "Enter a number:\n"; int number; cin >> number;

/* The maximum value is: 99 The mininum value is: -13 */#include <iostream>#include <fstream>#include <cctype>using namespace std;int main() { ifstream inStream; inStream.open("/Users/br/

运行结果： Enter the first of 19 or fewer digits number: 8837899188361066012 Enter the second of 19 or fewer digits number: 6832629488379649259 Integer overflow. Enter the first of 19 or fewer digits numb

#include <iostream>#include <string> // 用于C++ string类的操作#include <cctype> // 用于C字符串的拷贝、比较、追加 等，strcpy(), strcmp(), strcat()#include <cstdlib> // 用于exit(EXIT_FAILURE)及atoi即：alpha to interge 的缩写int m

运行结果： Enter the first money (dollars): 12.123 Invalid dollars and cents, negative values Enter the first money (dollars): 12.21 Enter the second money (dollars): 99.99 The sum is: $112.20altMoney.h#

/* 77.6669okwari7.6669 0okwarisunsfa6+5+788 7 hello!*/#include <iostream>#include <fstream>#include <iomanip>using namespace std;int main() { cout.width(3); cout << 7 << endl; c

运行结果： Enter the first money (dollars): 12.123 Invalid dollars and cents, negative values Enter the first money (dollars): 33.99 Enter the second money (dollars): 35.33 The sum is: $69.32AltMoney.h#i

* Enter the number fo seconds:31600000 * 31600000-seconds = 365 days, 17 hours, 46 minutes, 40-seconds.

/*Enter a point's x- and y-coordinates: 100.5 25.5The point is in the triangle.Enter a point's x- and y-coordinates: 100.5 50.5The point is not in the triangle.*/import java.util.Scanner;public cl

manhattan.txt15.33 21.11 3.00 10.32mami.txt11.33 91.11 13.00 20.32#include <iostream>#include <fstream>using namespace std;int main() { ofstream mia;

程序说明:读入一个字符串，当字符串以.结尾且后面无任何数据时，停止读入。 读入后的字符串第一个字母必须是大写，且其余字母均为小写并过滤掉字符之间多余的空格。 运行结果： Enter a string: the Answer to life, the Universe, and everything IS 42. The answer to life, the universe, an

运行结果： Enter a string: My userID is john17 and my 4 digit pin is 1234 which is secret. My userID is john17 and my x digit pin is xxxx which is secret.replaceDigits.cpp#include <iostream>#include <stri

score.txtscore.txtXu Lisa 60 70 80 90 50 40 30 20 10 0Cheng Kevin 1 2 3 4 5 6 7 8 9 10Coco Li 11 22 33 44 55 66 77 88 99 100scoreAvg.txtXu Lisa 60 70 80 90 50 40 30 20 10 0 average: 45Cheng Kev

运行结果： Enter the number between 1 and 20: 5 ***** * * * * * * *****main.cpp#include <iostream>using namespace std;int main() { unsigned int row; cout << "Enter the number between 1 and

运行结果： Rolls 10 Jam 3 Tea 2 Enter you want to insert after the item: Tea Enter the insert item: Lisa Enter the insert count: 33 Rolls 10 Jam 3 Tea 2 Lisa 33 Do you want to continue <y/n

运行结果：a = 9000a + 100 = 9100a + 834 = 9834100 + a = 9100bye = 9000Overflow, Product is over than 4-digits.main.cpp#include <iostream>#include <stdexcept>#include "Huge.hpp"#include <string>int m

input.txtDoe John 555555555 333-333-3333 65000 Doe Jan 444444444 333-333-3333

注：二分查找法有个问题，一组数字中有多个一样的数值时，其返回的下标位置不一样是数组中第一次出现的线标位置。 运行结果： Enter digit numbers less than 10: -5 5 -5 33 -10 4 1 33 5 After sort: -10 -5 -5 1 4 5 5 33 33 Enter you want to find number: -5 Found it

注：如果要计算double类型的数据，只需将int改为double即可，如果只是计算整数值的N次方不建议用double类型， 因为double类型只能存储一个数的近似值，所以计算的结果部分时候会有误差。运行结果：Enter a number to the power of another such as (x n): 2 0The number 2 power of 0 is:

运行结果： Enter a name to find the corresponding phone number. Ash Williams The number is: 333-2323 Look up another name? (y/n) y Enter a name to find the corresponding phone number. lenove Name not

运行结果：Enter a string: The new operator takes a type for its argument..tnemugra sti rof epyt a sekat rotarepo wen ehTreverseString.cpp#include <iostream>#include <string>using namespace std;void revers

运行结果： Enter a string: For example, the input 4 words 1 a 3 e 1 f 1 h 1 i 1 l 1 m 1 n 1 o 2 p 1 r 2 t 1 u 1 xcountWords.cpp#include <iostream>#include <string>#include <cstring>using na

Enter numbers: 0 -3.45 -1 -66 100 32.32 54.12 9.99 8.09 -6.66 -3.44 -1 1.99 -1 0 9 p Ascending sort: -66 -6.66 -3.45 -3.44 -1 -1 -1 0 0 1.99 8.09

未加密文本.txtalmost loverYour fingertips against my skin The palm trees swaying in the wind in my chase You sang me Spanish lullabies The sweetest sadness in your eyes Clever trickI never wanna see you

追加字符串到指定文件

运行结果：int 最大值：2147483647int 最小值：-2147483648long 最大值：9223372036854775807long 最小值：-9223372036854775808long long 最大值：9223372036854775807long long 最小值：-9223372036854775808unsigned long long 最大值：184467

Enter two number: 2 3.5 5.6 2.1 11.89 44.6 0 23.09 2 和 3.5的调和平均数为：2.54545 5.6 和 2.1的调和平均数为：3.05455 11.89 和 44.6的调和平均数为：18.7748

#include <iostream>using namespace std;const int SLEN = 30;struct student { char fullname[SLEN]; char hobby[SLEN]; int ooplevel;};int getinfo(student pa[], int n);void display1(student

计算平均值并显示有几个数字大于平均值

计算Cleo需要多少年收益可赶超Daphne并显示各自的总投资额。

Enter a line: get out " get out " contains 7 characters 7 characters total Enter next line (empty line to quit):

ad: x = 10.5 , y = 6.5 ae: x = 4.44 , y = 2.11 ad + ae: x = 14.94 , y = 8.61 x = 0 , y = 0 x = 0 , y = 0

getString.cpp#include <iostream>#include <fstream>void getInputStream(std::ifstream& outFile);int main() { using std::cout; using std::string; std::ifstream outFile; string s; getIn

运行结果：-12 -12 -12 -12 0 0 1 1 1 2 2 3 3 3 4 4 4 4 3 3 3 2 2 1 1 1 0 0 -12 -12 -12 -12 insertionSort.cpp#include <iostream>using namespace std;void ascendingSort(int ar[]

/* Enter 2 values for X and Y separated by space, The press <enter>: 2.4 4.85 2.4 to the power of 4.85 is = 69.8272 Enter 2 values for X and Y separated by space, The press <enter>: 3 4 3 to th

Enter a number: 5 5!= 120 Enter a number: 6 6!= 720

Enter a double number: 1 2 3 4 5 q Input 5 number to d[dSzie]. all d[0] ~ d[10]: 1 2 3 4 5 all d[0] ~ d[10]: 5 4 3 2 1

/* Enter value #1: 10000 Enter value #2: 8000 Enter value #3: 2000 Enter value #4: 10000 Enter value #5: -9 Value #1: 10000 Value #2: 8000 Value #3: 2000 Value #4: 10000 Enter ravalue #1: 0.8