Bull Math（高精度）

最新推荐文章于 2022-03-06 13:27:02 发布

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#高精度 #指针 #内存 #malloc #cstring

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本文介绍了一种不依赖特殊库函数的大数乘法算法实现。通过字符串操作和基本的数学原理，该方法能够准确计算两个大整数的乘积，并以常规格式输出结果。

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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

代码借鉴的别人的，有待改善，可以不用指针，这题是两个大数的乘法。

AC代码

#include<iostream>
#include <stdlib.h>
#include<stdio.h>
#include<cstring>
using namespace std;
#define MAX 45
char mul1[MAX],mul2[MAX];
char ans[2 * MAX];

void mult(char* a, char* b, char* c)
{
    int i, j, ca, cb, *s;
    ca = strlen(a);
    cb = strlen(b);
    s=(int*)malloc(sizeof(int)*(ca+cb));                   //给s分配长度为ca+cb的内存块
    for(i = 0; i < ca + cb; i++)
        s[i] = 0;
    for(i=0; i<ca; i++)
        for(j = 0; j < cb; j++)
            s[i + j + 1] += (a[i] - '0') * (b[j] - '0');
    for(i = ca + cb - 1; i >= 0; i--)
        if(s[i] >= 10)
        {
            s[i-1] += s[i]/10;
            s[i] %= 10;
        }

    i=0;
    while(s[i] == 0)
        i++;
    for (j = 0; i < ca + cb; i++, j++)
        c[j] = s[i] + '0';
    c[j] = '\0';
    free(s);
}
int main()
{
     scanf("%s", mul1);
     scanf("%s", mul2);
     mult(mul1, mul2, ans);
     printf("%s\n", ans);
     return 0;
}