大数乘法（C++）

发布时间：2020-12-14 02:40:15 所属栏目：大数据来源：网络整理

导读：题目：POJ 2398 Bull Math Time Limit: ?1000MS ? Memory Limit: ?65536K Total Submissions: ?13410 ? Accepted: ?6903 Description Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise

题目：POJ 2398

Bull Math

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?13410	?	Accepted:?6903

Description

Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros).?

FJ asks that you do this yourself; don't use a special library function for the multiplication.

Input

* Lines 1..2: Each line contains a single decimal number.

Output

* Line 1: The exact product of the two input lines

Sample Input

22222222221111
2222222222

Sample Output

12345679011110987654321

Source

USACO 2004 November

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
using namespace std;

const int maxn = 100;
void reverse(char a[])
{
    int len = strlen(a);
    for(int i = 0 ; i < len / 2; i++)
    {
        int temp = a[i];
        a[i] = a[len - i - 1];
        a[len - i -1] = temp;
    }
}
int main()
{
    char a[maxn],b[maxn];
    int t[100] = {0};
    //printf("Please enter 2 numbers: ");
    scanf("%s%s",a,b);
    reverse(a);
    reverse(b);
    if(strcmp(a,"0")==0||strcmp(b,"0")==0)
        cout<<"0"<<endl;
    else
    {
        int i,j;
        for(i = 0; i <strlen(b); i++)
        {
            int cnt = 0;
            for(j = 0; j < strlen(a); j++)
            {
                int temp = (b[i] - '0') * (a[j] - '0');
                int tt= t[i+j] + temp + cnt;
                t[j+i] = tt % 10;
                cnt = tt / 10;
            }
            while(cnt != 0)
            {
                t[j+i] = cnt % 10;
                cnt = cnt / 10;
                j++;
            }
        }
        for(int k = i + j - 2; k >= 0; k--)
        {
            cout<<t[k];
        }
    }
    return 0;
}

（编辑：李大同）

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