实验 10.31

1. 实验 10.31

2. Lab4-Ex2 A right triangle can have sides that are all integers. A set of three integer values for the sides of a right triangle is called a Pythagorean triple. These three sides must satisfy the relationship that the sum of the squares of two of the sides is equal to the square of the hypotenuse. Find all Pythagorean triples for side1, side2 and hypotenuse all no larger than 500.

3. int main(){ int count = 0; // number of triples found long int hypotenuseSquared; // hypotenuse squared long int sidesSquared; // sum of squares of sides cout << "Side 1\tSide 2\tSide3" << endl; // side1 values range from 1 to 500 for ( long side1 = 1; side1 <= 500; side1++ ) { for ( long side2 = side1; side2 <= 500; side2++ ) { for ( long hypotenuse = side2; hypotenuse <= 500; hypotenuse++ ) { hypotenuseSquared = hypotenuse * hypotenuse; sidesSquared = side1 * side1 + side2 * side2; if ( hypotenuseSquared == sidesSquared ) { cout << side1 << '\t' << side2 << '\t' << hypotenuse << '\n'; count++; // update count break; } // end if } // end for } // end for } // end for cout << "A total of " << count << " triples were found." << endl; return 0; // indicate successful termination } // end main

4. Lab5_Ex1 • 分别用三个函数实现绘制正方形(square)、菱形(diamond)、三角形(triangle)，在main函数中用循环结构嵌套switch结构实现对各种图形函数的调用测试。

6. #include <iostream> #include <cmath> using namespace std; void square( int, char ); // function prototype void diamond( int, char ); // function prototype void triangle ( int, char ); // function prototype int main(){ int size; char character; int shape; char answer; do{ cout << "Choose the shape to graph" << endl<< "1 for square" << endl << "2 for diamond" << endl<< "3 for triangle, " << endl << "4 for circle: " << endl << "? "; cin >> shape;cout << "Enter a character and size: "; cin >> character >> size;cout << '\n'; switch( shape ){ case 1: square( size, character );break; case 2: diamond( size, character );break; case 3: triangle( size, character );break; default: square( size, character );break; } // end switch cout<<"do you want to continue? ( y or n)"<<endl; cin>>answer; }while(answer=='y'); return 0; // indicate successful termination } // end main

7. // square displays solid square of fillCharacter with specified side void square( int side, char fillCharacter ) { // loop side times for number of rows for ( int row = 1; row <= side; row++ ) { // loop side times for number of columns for ( int col = 1; col <= side; col++ ) cout << fillCharacter; cout << endl; } // end for cout << endl; } // end function square

8. void diamond( int size, char fillCharacter ){ int rows; // top half for ( rows =size%2; rows <= size - 2; rows += 2 ){ // print preceding spaces for ( int space = ( size - rows) / 2; space > 0; space-- ) cout << ' '; // print characters for ( int character = 1; character <= rows; character++ ) cout << fillCharacter; cout << endl; } // end for // bottom half for ( rows = size; rows >= 0; rows -= 2 ){ // print preceding spaces for ( int space = ( size - rows ) / 2; space > 0; space-- ) cout << ' '; // print characters for ( int character = 1; character <= rows; character++ ) cout << fillCharacter; cout << endl; } // end for cout << endl; } // end function diamond

9. // triangle displays solid triangle of fillCharacter // with specified number of rows void triangle( int size, char fillCharacter ) { // display triangle for ( int row = 1; row <= size; row++ ) { // print preceding spaces for ( int space = size - row; space > 0; space-- ) cout << ' '; // print characters for ( int character = 1; character < 2 * row; character++ ) cout << fillCharacter; cout << endl; } // end for cout << endl; } // end function triangle

10. Lab6_Ex2 • 一个完全数是这样的一个整数，它的所有因子（包含1但不包含该数本身）的和恰好等于该数本身。举个例子，6是一个完全数，因为6=1+2+3。编写一个函数perfect(bool类型返回值)，用它来判断该函数的形式参数是否为一个完全数。在程序中调用该函数，找出1至1000这个范围内所有的完全数并打印出来。

11. #include <iostream> using std::cout; using std::endl; bool isPerfect( int ); // function prototype void printSum( int ); // function prototype int main() { cout << "Perfect integers between 1 and 1000:" << endl; // loop from 2 to 1000 for ( int j = 2; j <= 1000; j++ ) { // if current integer is perfect if ( isPerfect( j ) ) printSum( j ); // print it as sum of factors } // end for cout << endl; return 0; // indicate successful termination } // end main

12. // isPerfect returns true if value is perfect integer, // i.e., if value is equal to sum of its factors bool isPerfect( int value ) { int factorSum = 1; // current sum of factors // loop through possible factor values for ( int i = 2; i <= value / 2; i++ ) { // if i is factor if ( value % i == 0 ) factorSum += i; // add to sum } // end for // return true if value is equal to sum of factors return factorSum == value ? true : false; } // end function isPerfect

13. void printSum( int value ) { cout << value << " = 1"; // loop through possible factor values for ( int i = 2; i <= value / 2; i++ ) { // if i is factor if ( value % i == 0 ) cout << " + " << i; // display as part of sum } // end for cout << endl; } // end function printSum

14. Lab6_Ex3 （质数问题）一个质数是这样的一个整数，它只能被1和该数本身整除。举个例子，2，3，4，7是质数，而4，6，8，9不是质数。 1.编写一个函数来判断一个数是否为质数。 2.在程序中调用这个函数，找出2至10000这个范围内所有的质数并且打印出来。在你确认找出所有质数之前你测试了多少次呢？ 3.最初，你可能认为判断一个数是否为质数的测试次数上限为n/2，但实际你需要的测试次数的上限只是n的平方根。这是为什么呢？重新编写这个程序，分别在这两种方式下运行，并观察出性能的改进。

15. #include <iostream> #include <iomanip> #include <cmath> using namespace std; bool isPrime( int ); // function prototype int main(){ int count = 1; // total number of primes found cout << "The prime numbers from 1 to 10000 are:" << endl; cout << setw( 6 ) << 2; // 2 is only even prime for ( int loop = 3; loop < 10000; loop += 2 ) { if ( isPrime( loop ) ) { count++; cout << setw( 6 ) << loop; if ( count % 10 == 0 )// new line after 10 values displayed cout << '\n'; } // end if } // end for cout << endl << "Total of " << count << " prime numbers between 1 and 10000." << endl; return 0; // indicate successful termination } // end main

16. // isPrime returns true if n is prime bool isPrime( int n ) { // loop through possible factors for ( int i = 2; i <= n / 2; i++ ) //for ( int i = 2; i <= static_cast< int > ( sqrt( static_cast< double > ( n ) ) ); i++ ) { // if factor found, not prime if ( n % i == 0 ) return false; } // end for return true; } // end function prime

17. Lab6_Ex4 • 两个整型数的最大公约数(The greatest common divisor, GCD)定义为最大的整除这两个整数的数。编写gcd函数，返回两个整型数的最大公约数。

18. #include <iostream> using std::cin; using std::cout; using std::endl; int gcd( int, int ); // function prototype int main(){ int a; // first number int b; // second number cout << "Enter two integers: "; cin >> a >> b; cout << "The greatest common divisor of " << a << " and " << b << " is "; // find greatest common divisor of a and b cout << gcd( a, b ) << endl; return 0; // indicate successful termination } // end main

19. int gcd( int x, int y ) { int greatest = 1; // loop from 2 to smaller of x and y for ( int i = 2; i <= ( ( x < y ) ? x: y ); i++ ) { // if current i divides both x and y if ( x % i == 0 && y % i == 0 ) greatest = i; } // end for return greatest; } // end function gcd