Questin：

There is an array A[N] of N numbers. You have to compose an array Output[N] such that Output[i] will be equal to multiplication of all the elements of A[N] except A[i]. Solve it without division operator and in O(n).

For example Output[0] will be multiplication of A[1] to A[N-1] and Output[1] will be multiplication of A[0] and from A[2] to A[N-1].

Example:

A: {4, 3, 2, 1, 2}

OUTPUT: {12, 16, 24, 48, 24}

思路：

可以使用迭代累计。把output[i]=a[i]左边的乘积 x a[i]右边的乘积，所以，我们可以分两个循环，第一次先把A[i]左边的乘积放在Output[i]中，第二次把A[i]右边的乘积算出来；也可以直接放在一个循环中，只不过需要同时计算左边的乘积和右边的乘积。（当然也可分开计算A[i]左边和右边的数据，这样容易理解！最后将左边和右边的相乘即可）

代码如下：

void Multiplication_Array(int A[], int OUTPUT[], int n) {  int left = 1;  int right = 1;  for (int i = 0; i < n; i++)      OUTPUT[i] = 1;  for (int i = 0; i < n; i++) {      OUTPUT[i] *= left;      OUTPUT[n - 1 - i] *= right;      left *= A[i];      right *= A[n - 1 - i];  }}

void Mutiplication_Array2() {

    int *X = new int[n];

    int *Y = new int[n];

    // Create X

    X[0] = 1;

    for(int i = 1; i < n; i++){

        X[i] = X[i-1] * A[i-1];

    }

    // Create Y

    Y[n-1] = 1;

    for(int i = n-2; i >= 0; i--){

        Y[i] = Y[i+1] * A[i+1];

    }

    // Create Out

    for(int i = 0; i < n; i++){

        out[i] = X[i] * Y[i];

    }

    // Delete X and Y

    delete[] X;

    delete[] Y;

}