"Sword Point offer": [52] Building a product array

title: Given an array of a[0,1,2...n-1], build an array b[0,1,2,... N-1], so that the element in B is b[i]=a[0]*a[1]*...*a[i-1]*a[i+1]*...*a[n-1], you cannot use division.
Solution Ideas:Through the positive and negative two times to seek b[i];
Zheng: b[i]=a[0]* ... A[I-1];
Inverse: temp = b[n]* ... B[I+1];
Finally, the results can be obtained by b[i]*temp.
Take a[0,1,2,3,4], b[0,1,2,3,4] For example:
Step Analysis:
First step: b[0] = 1;
Step two: b[1] = b[0] * A[0] = a[0]
Step three: b[2] = b[1] * A[1] = a[0] * a[1];
Fourth step: b[3] = b[2] * a[2] = a[0] * a[1] * a[2];
Fifth step: b[4] = b[3] * A[3] = a[0] * a[1] * a[2] * a[3];
Then for a second for loop
The first step:
Temp *= a[4] = 1*a[4];
B[3] = b[3] * temp = a[0] * a[1] * a[2] * a[4];
Step Two:
Temp *= a[3] = a[4] * A[3];
B[2] = b[2] * temp = a[0] * a[1] * a[4] * a[3];
Step Three:
Temp *= a[2] = a[4] * a[3] * a[2];
B[1] = b[1] * temp = a[0] * a[4] * a[3] * a[2];
Fourth Step:
Temp *= a[1] = a[4] * a[3] * a[2] * a[1];
B[0] = b[0] * temp = a[4] * a[3] * a[2] * a[1];
The specific implementation code is as follows:

#include <iostream>using namespace std; #include <vector>vector<int> array1;vector<int> Array2 (5); void Multiply (const vector<int> &array1,vector<int> &array2) {int length1=array1.size ( ); int length2=array2.size (); if (length1==length2 && length2>1) {array2[0]=1;for (int i=1;i<length1;i++) {array2[i]=array2[i-1]*array1[i-1];} Double temp=1;for (int i=length1-2;i>=0;i--) {temp*=array1[i+1];array2[i]*=temp;}}} int main () {for (int i=0;i<5;i++) array1.push_back (i+1); multiply (array1,array2); Vector<int>::iterator it1; Vector<int>::iterator it2;cout<< "A array is:"; for (It1=array1.begin (); It1!=array1.end (); it1++) cout<< *it1<< "";cout<<endl;cout<< "b array is:"; for (It2=array2.begin (); It2!=array2.end (); it2++) cout< <*it2<< ""; Cout<<endl;system ("pause"); return 0;}

The main use of intermediate result variables. Operation Result:

"Sword Point offer": [52] Building a product array