Bull Math（高精度）

最新推荐文章于 2021-01-26 18:02:50 发布

userluoxuan

最新推荐文章于 2021-01-26 18:02:50 发布

阅读量503

点赞数

CC 4.0 BY-SA版权

分类专栏：解题报告 ===高精度=== 文章标签：高精度

本文链接：https://blog.youkuaiyun.com/userluoxuan/article/details/38316963

解题报告同时被 2 个专栏收录

71 篇文章

订阅专栏

===高精度===

8 篇文章

订阅专栏

本文介绍了一种不依赖特殊库函数的手动实现大数乘法的方法。通过读取两个不超过40位的正整数并计算它们的乘积，帮助验证大数运算的准确性。文章提供了详细的解题思路及AC代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

11111111111111
1111111111

Sample Output

12345679011110987654321

解题思路：

普通的高精度乘法题。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 100;
int main()
{
    int num_1[maxn], num_2[maxn], length_1, length_2, result[maxn * 2];
    char str_1[maxn], str_2[maxn];
    scanf("%s%s", str_1, str_2);
    memset(num_1, 0, sizeof(num_1));
    memset(num_2, 0, sizeof(num_2));
    memset(result,0,sizeof(result));
    length_1 = strlen(str_1);
    length_2 = strlen(str_2);
    for(int i = 0; i < length_1; i++)
    {
        num_1[i] = str_1[length_1-i-1] - '0';
    }
    for(int i = 0; i < length_2; i++)
    {
        num_2[i] = str_2[length_2-i-1] - '0';
    }
    for(int i = 0; i < length_1; i++)
        for(int j = 0; j < length_2; j++)
        {
            result[i + j] += num_1[i] * num_2[j];
        }
    for(int i = 0; i < maxn * 2; i++)
    {
        if(result[i] >= 10)
        {
            result[i + 1] += result[i]/10;
            result[i] %= 10;
        }
    }
    bool isBegin = false;
    for(int i = maxn * 2 - 1; i >= 0; i--)
    {
        if(isBegin)
            printf("%d",result[i]);
        else if(result[i])
        {
            printf("%d",result[i]);
            isBegin = true;
        }
    }
    if(!isBegin)
        printf("0");
    return 0;
}