An array a [n] is known to construct an interesting Algorithm for Array B [N ].

Given an array a [n], we want to construct array B [N], where B [I] = A [0] * A [1] *... * A [N-1]/A [I].
The construction process must meet the following requirements:
1. Division is not used;
2,O (1)Spatial complexity andO (N)Time complexity;
3. In addition to traversing the counters used by a [n] B [N], no new variables (including stack temporary variables, space-based and Global static variables) are used );

 

/*************************************** * ****** // * Specify an array a [n], we want to construct the array B [N], where B [I] = A [0] * A [1] *... * A [N-1]/A [I]. Division is not allowed in the construction process: 1; 2 requires O (1) space complexity and O (n) time complexity; 3. In addition to traversing the counter and a [n] B [N], new variables (including stack temporary variables, space-and Global static variables) cannot be used ); *//************************************* * *******/void makearray (int A [], int B [], int Len) {int I, j; B [0] = 1; for (I = 1; I <Len; I ++) {B [I] = B [I-1] * A [I-1]; // tired ride a [0] * A [1]... A [I-1]} B [0] = A [len-1]; for (j = len-2; j> 0; j --) {B [J] * = B [0]; B [0] * = A [J];}

This question is very common. After reading it, I analyzed it. If in B [N], B [I] = A [0] * A [1] *... * A [N-1]/A [I], then we can consider separately, the implementation code uses two loops, the first loop first implementation,

B1 [I] = A [0] a [1] A [2]… A [I-1], the next cycle from the back to achieve B2 [I] = A [len-1] a [len-2]... A [I + 1], in the second cycle, B [I] = b1 [I] * B2 [I] is implemented due to the constant changes of B [0]. therefore, the question requirement is met. Because the variable cannot be defined in the question, B [0] can be used multiple times.

An array a [n] is known to construct an interesting Algorithm for Array B [N ].