Write and run a program that reverses the digits of a given positive integer.

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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: main0 l long m, d, n = 0; tout -c-c "Enter a positive integer: '; tin >> m; while (m > 0) { d = m % 10; // d will be the right-most digit of m m /= 10; // then remove that digit from m n = lO*n + d; // and append that digit to n > tout << "The reverse is I' KC n <C endl; Enter a positive integer: I223456 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.

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ITERATION AND FLOATING TYPES

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Rewrite the for loop in Example 3.6, using the conditional expression operator in place of the if statements.

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The conditional expression ( n c min n : min) evaluates to n if n < min, and it evaluates to min otherwise. So assigning that value to min is equivalent to the first line of the for loop in the example. Similarly, the assignment max = ( n > max n : min ) is equivalent to the second line in the other for loop.

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(min=max=n; n>O;) { min = (n c min n : min); max = (n > max n : min); tin >> n;

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// min and max are the smallest // and largest of the n that // have been read so far

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Note that in this version we did not use an equivalent to the el se i f . 3.19

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Implement the Euclidean Algorithm for finding the greatest common divisor of two given pos\ itive integers.

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The Euclidean Algorithm transforms a pair of positive integers (m, n) into a pair (d, 0) by repeatedly dividing the larger integer by the smaller integer and replacing the larger with the remainder. When the remainder is 0, the other integer in the pair will be the greatest common divisor of the original pair (and of all the intermediate pairs). For example, if m is 532 and y2 is 112, then the Euclidean Algorithm reduces the pair (532,112) to CWO) by

(532,112) + (112,84) + (84,28) + (28,0)

So 28 is the greatest common divisor of 532 and 112. This result can be verified from the facts that 532 = 28.19 a n d 112 = 28.8. The reason that the Euclidean Algorithm works is that each pair in the sequence has the same set of divisors, which are precisely the factors of the greatest common divisor. In the example above, that common set of divisors is { 1, 2, 4, 7, 14, 28). The reason that this set of divisors is invariant under the reduction process is that when m = n-q + r, a number is a common divisor of m and y1 if and only if it is a common divisor of y2 and r.

main0 // begin scope of main0 { int m, n, r; tout << "Enter two positive integers: I'; tin >> m >> n; if (m c n) { int temp = m; m = n; n = temp; } // make m >= n tout << "The g.c.d. of ' CC m CC ' and ' CC n CC' ' is "; while (n > 0) { r =m%n; m = n; n = r; tout CC m CC endl;

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