一个长度为n的数组a[0],a[1],...,a[n-1]。现在更新数组的每个元素,
即a[0]变为a[1]到a[n-1]的积,
a[1]变为a[0]和a[2]到a[n-1]的积,
...
,a[n-1]为a[0]到a[n-2]的积。
程序要求:要求具有线性复杂度,不能使用除法运算符。
思路是思想跟这个一样,用两个数组b、c
b[i] = a[0] *... * a[i - 1],
c[i] = a[i + 1] *...*a[n] ,时间复杂度为O(2n)
最后a[i] = b[i]*c[i]。
#include<iostream>
#include <stdio.h>
using namespace std;
int main()
{
int n =5;
int * a = new int[n];
int * b = new int[n];
int * c = new int[n];
for (int i = 0; i< n; i++)
{
a[i]=i+1;
}
b[0]=1;
b[1]=a[0];
for (int i =2;i<n;i++)
{
b[i]=b[i-1]*a[i-1];
}
c[n-1]=1;
c[n-2]=a[n-1];
for (int i =n-3;i>=0;i--)
{
c[i]=c[i+1]*a[i+1];
}
for (int i=0;i<n;i++)
{
a[i]=b[i]*c[i];
}
for(int i=0;i<n;i++)
{
cout<<a[i]<<" "<<endl;
}
delete [] a;
delete [] b;
delete [] c;
return 0;
}
C:\MinGW\01-笔试面试题>a.exe
120
60
40
30
24