Two arrays A [n], B [N], where each element value of a [n] is known, and B [I] is assigned a value, B [I] = A [0] * A [1] * A [2]… * A [N-1]/A [I]

[Problem]

1. Division is not required.

Two arrays A [n], B [N], where each element value of a [n] is known, and B [I] is assigned a value, B [I] = A [0] * A [1] * A [2]... * A [N-1]/A [I]; requirements:

• 1. Division operations are not allowed.
• 2. Except for the cyclic Count value, a [n], B [N], no other variables (including local variables and global variables) are allowed)
• 3. time complexity O (N) and space complexity O (1 ).

[Analysis]

Tip: Question B [I] = A [0] * A [1] * A [2]... * A [N-1]/A [I], equivalent to a [0] * A [1] * A [2] * A [3]... A [I-1] * A [I + 1] .. * A [N-1] is equivalent to removing the product of the current element a [I], all other elements (the left part of a [I], and the right part of a [I.

Note left [I] = semi A [K], (k = 1... i-1); Right = clerk a [K], (k = I + 1... n) according to the description of B [I] = left [I] * right [I], initialize to 1 for each B [I, left [I] And right [I] can be multiplied by two times separately, that is, for the cyclic variable I = 1... n, B [I] = left [I]; B [n-I] = right [n-I]. The computation can be completed after the loop is completed.

[Code]

#include <stdio.h>#include <stdlib.h>void Multiplication(int *a, int *b, int length){int i;b[0] = 1;for (i = 1; i < length; i ++) {b[0] *= a[i - 1];b[i] = b[0];}b[0] = 1;for (i = length - 2; i > 0; i--) {b[0] *= a[i + 1];b[i] *= b[0];}b[0] = a[1] * b[0];}int main(void) {int i;int a[] = {2, 2, 3, 4, 5};int b[5];int length = sizeof(a) / sizeof(int);Multiplication(a, b, length);for (i = 0; i < length; i++)printf("%d\t", b[i]);return 0;}