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sum += i*i; cout << "The sum of the first " << n << " squares is " << sum << endl; } Enter a positive integer: 6 The sum of the first 6 squares is 91 This program uses a do..while loop to compute the sum of the first n squares, where n is input: int main() { int n; cout << "Enter a positive integer: "; cin >> n; int sum=0, i=1; do { sum += i*i; } while (i++ < n); cout << "The sum of the first " << n << " squares is " << sum << endl; } Enter a positive integer: 6 The sum of the first 6 squares is 91 This program directly implements the quotient operator / and the remainder operator % for the division of positive integers. The algorithm used here, applied to the fraction n/d, repeatedly subtracts the d from the n until n is less than d. At that point, the value of n will be the remainder, and the number q of iterations required to reach it will be the quotient: int main() { int n, d, q, r; cout << "Enter numerator: "; cin >> n; cout << "Enter denominator: "; cin >> d; for (q = 0, r = n; r >= d; q++) r -= d; cout << n << " / " << d << " = " << q << endl; cout << n << " % " << d << " = " << r << endl; cout << "(" << q << ")(" << d << ") + (" << r << ") = " << n << endl; } Enter numerator: 30 Enter denominator: 7 30 / 7 = 4 30 % 7 = 2 (4)(7) + (2) = 30 This run iterated 4 times: 30 7 = 23, 23 7 = 16, 16 7 = 9, and 9 7 = 2. So the quotient is 4, and the remainder is 2. Note that this relationship must always be true for integer division: (quotient)(denominator) + (remainder) = numerator The trick here is to strip off the digits one at a time from the given integer and accumulate them in reverse in another integer: int main() { long m, d, n = 0; cout << "Enter a positive integer: "; cin >> m;
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while (m > 0) { d = m % 10; // d will be the right-most digit of m m /= 10; // then remove that digit from m n = 10*n + d; // and append that digit to n } cout << "The reverse is " << n << endl; } Enter a positive integer: 123456 The reverse is 654321 In this run, m begins with the value 123,456. In the first iteration of the loop, d is assigned the digit 6, m is reduced to 12,345, and n is increased to 6. On the second iteration, d is assigned the digit 5, m is reduced to 1,234, and n is increased to 65. On the third iteration, d is assigned the digit 4, m is reduced to 123, and n is increased to 654. This continues until, on the sixth iteration, d is assigned the digit 1, m is reduced to 0, and n is increased to 654,321. This implements the Babylonian Algorithm: #include <cmath> // defines the fabs() function #include <iostream> using namespace std; int main() { const double TOLERANCE = 5e-8; double x = 2.0; while (fabs(x*x - 2.0) > TOLERANCE) { cout << x << endl; x = (x + 2.0/x)/2.0; // average of x and 2/x } cout << "x = " << x << ", x*x = " << x*x << endl; } 2 1.5 1.41667 1.41422 x = 1.41421, x*x = 2 We use a tolerance of 5e-8 (= 0.00000005) to ensure accuracy to 7 decimal places. The fabs() function (for floating-point absolute value ), defined in the <cmath> header file, returns the absolute value of the expression passed to it. So the loop continues until x*x is within the given tolerance of 2. This program finds the integer square root of a given number. This method uses an exhaustive algorithm to find all the positive integers whose square is less than or equal to the given number: int main() { float x; cout << "Enter a positive number: "; cin >> x; int n = 1; while (n*n <= x) ++n; cout << "The integer square root of " << x << " is " << n-1 << endl; } Enter a positive number: 1234.56 The integer square root of 1234.56 is 35
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