使用数组计算大量乘积?

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我有一个项目(用于学校)，我绝对不能使用任何外部库，因此不能使用任何大数库，我需要得到2个(非常大数)的乘积.所以我想我会为此写我自己的代码，但是我似乎无法获得传递一位数的乘法.

I have a project (for school) and I absolutely cant use any external libraries hence cannot use any big numbers library and I need to get the product of 2 (very) large numbers. So I thought I'll actually write my own code for it but I cant seem to get pass single digit multiplications.

到目前为止，我的工作方式是我有一个字符'a'数组.病态将其每个数字与另一个数字相乘(因为任何乘法都不能超过81，即9 * 9).但是我似乎无法弄清楚如何将两个数组彼此相乘.

How I've done it so far is I have an array of chars 'a'. And Ill multiply each of its digits with the other number (since no multiplication can go beyond 81 ie, 9*9). But I cant seem to figure out how Ill multiply two arrays with each other.

如

int a[] = {1,2,3}; int b[] = {4,5,6}; int r[200]; // To store result of 123x456. After processing should have value 56088

到目前为止，这里有我的代码...

Heres my code so far...

#include <iostream> using namespace std; void reverseArray(int array[], int n) { int t; for(int i=0;i<n/2;i++) { t = array[i]; array[i] = array[n-i-1]; array[n-i-1] = t; } } int main() { int A[] = {1,2,6,6,7,7,8,8,8,8,8,8,8,8,8,8}; int s = sizeof(A)/sizeof(int); int n = s-1; int R[50]; int x = 2; int rem = 0; for(int i=0; i<s; i++) { R[i] = (A[n-i] * x) % 10; R[i] += (rem != 0) ? rem:0; rem = (A[n-i] * x) / 10; } reverseArray(R, s); for(int i=0; i<s; i++) cout<<R[i]; // Gives 2533557777777776 }

我还找到了一个类似的程序此处，用于计算非常大的阶乘.但是我似乎无法充分理解代码以适应自己的需求.

I also found a similar program here which calculates factorials of very large numbers. But I cant seem to understand the code enough to change it to my needs.

很抱歉，这个问题有点粗略.

Sorry if the question is a little sketchy.

谢谢.

推荐答案

我已经用Java完成了，在这里我将数字N1和N2进行了处理，并创建了一个大小为1000的数组.这，N1 = 12，N2 = 1234.对于N1 = 12，temp = N1％10 = 2，现在将此数字从右到左与数字N2相乘，并将结果存储到从i = 0开始的数组中，类似于N1的其余数字.数组将存储结果，但顺序相反.看看这个链接. ideone/UbG9dW#view_edit_box

I have done in java, Here I am taking to numbers N1 and N2, And I have create an array of size 1000. Lets take an example How to solve this, N1=12, N2=1234. For N1=12, temp=N1%10=2, Now Multiply this digit with digit N2 from right to Left and store the result into array starting from i=0, similarly for rest digit of N1. The array will store the result but in reverse order. Have a looking on this link. ideone/UbG9dW#view_edit_box

//Product of two very large number import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; class Solution { public static void main(String[] args) { Scanner scan=new Scanner(System.in); int N1=scan.nextInt(); int N2=scan.nextInt(); //int N=scan.nextInt(); int [] array=new int[1000]; Arrays.fill(array,0); int size=multiply(N1,N2,array); for(int i=size-1;i>=0;i--){ System.out.print(array[i]); } } public static int multiply(int N1, int N2, int [] result){ int a=N1; int b=N2; int count=0, carry=0; int i=0; int max=0; if(a==0||b==0) return 1; while(a>0){ int temp1=a%10; a=a/10; i=0; while(b>0){ int temp2=b%10; b=b/10; int product=result[count+i]+temp1*temp2+carry; result[count+i]=product%10; carry=product/10; i++; //System.out.println("ii="+i); } while(carry>0){ result[count+i]=carry%10; carry=carry/10; i++; //System.out.println("iiii="+i); } count++; b=N2; } //System.out.println("i="+i); return i+count-1; } }