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a[],

for (int i = 0; i c n; i++) tout << tl 'I cc a[i]; tout CC endl;

void print(const float a[], int index[], const int n) -t for (int i = 0; i c n; i++) tout cc l ' CC a[index[i]]; tout CC endl; >

The only modification needed to the Bubble Sort is to enclose each index with index [ . . . ] . So j is replaced with index [ j ] , and j +l is replaced with index [ j +l] . The effect is to leave the array a unchanged while moving the elements of the index array instead. Note that we have two overloaded print ( ) function: one to print the array directly, and the other

to print it indirectly using an index array. This allow us to check that the original array a is left unchanged by the indirect sort. 5.13

Implement the Sieve of Eratosthenes to find prime numbers. Set up an array prime [n] of ints, set a[O] = a[ I] = 0 (0 and 1 are not primes), and set a [ 2 ] through a [n- 11 to 1. Then for each i from 3 to n-l, set a [i] = 0 if i is divisible by 2 (i.e., i%2 == 0). Then for each i from4to n-1,set a[;] = 0 if i is divisible by 3. Repeat this process for each possible divisor from 2 to n / 2. When finished, all the is for which a [ i ] still equals 1 are the prime numbers. They are the numbers that have fallen through the sieve.

The test driver initializes the prime array with 1000 zeros. Then after invoking the sieve ( ) function, it prints those index numbers i forwhich prime[i] - - 1:

const int size = 500; void sieve(int prime[], const int n); main0 int prime[size] = (0); sieve(prime,size); for (int i = 0; i c size; i++) { if (prime[i]) tout CC i CC ' "; if ((i+l) % 50 == 0) tout CC endl; tout CC endl;

// Sets prime[i] = 1 if and only if i is prime: void sieve(int prime[], const int n) .i for (int i = 2; i c n; i++) prime[i] = 1; // assume all i > 1 are prime for (int p = 2; p C= n/2; p++) { for (int m = 2*p; m c n; m += p) prime[m] = 0; // no multiple of p is prime while (!prime[p]) // advance p to next prime ++p; 1 1

The sieve ( ) function initially sets prime [ i ] to 1 for each i > 2. Then it resets prime [ i ] to 0 again for every multiple m of a prime p.

5.14 Write and test the function void reverse(float a[], int n)

This function reverses the array, so that its last element becomes its first, its second-to-last element becomes its second, etc. Note that this is different from Example 5.1 which does not require the movement of any elements in the array.

This solution simply swaps each of the first n/ 2 second half of the array: elements with the corresponding element in the

void print(const float [I, const int); void reverse(float [I, const int); main0 1 float a[81 = { 8 8 . 8 , 4 4 . 4 , 77.7, 11.1, 33.3, 99.9, 66.6, 22.2); print(a, 8); reverse(a, 8); print(a, 8);

void reverse(float a[], const int n) float temp; for (int i = 0; i c n/2; i++) temp = a[i]; = a[n-i-l]; a[il a[n-i-l] = temp; 1

5.15 Write and test a function that implements the Per$iect Shufle of a one-dimensional array with an even number of elements. For example, it would replace {11,22,33,44,55,66,77,88} with {11,55,22,66,33,77,44,88X