CUIT 2016 新生訓練題第一週 C - Bull Math

C - Bull Math

 

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

題解：最簡單的高精度乘法模板題目。

#include<stdio.h>
#include<string.h>
using namespace std;

char str1[45],str2[45];
int num1[45],num2[45];
int res[100];

int main()
{
    while(~scanf("%s%s",str1,str2))
    {
        /*重置部分*/
        memset(res,0,sizeof(res));
        memset(num1,0,sizeof(num1));
        memset(num2,0,sizeof(num2));

        /*讀入與轉化部分*/
        int len1=strlen(str1);
        int len2=strlen(str2);
        for(int i=len1-1; i>=0; i--) num1[len1-1-i]=str1[i]-'0';
        for(int j=len2-1; j>=0; j--) num2[len2-1-j]=str2[j]-'0';

        /*乘法運算部分*/
        for(int i=0; i<len1; i++)
        {
            for(int j=0; j<len2; j++)
            {
                res[i+j]+=num1[i]*num2[j];
            }
        }

	/*進位部分*/
        int length=len1+len2-1;
        for(int i=0;i<length;i++)
        {
            if(res[i]>9)
            {
                res[i+1]+=res[i]/10;
                res[i]%=10;
                if(i+1==length)
                {
                    length++;
                }
            }
        }

s	/*輸出部分*/
        for(int i=length-1; i>=0; i--)
        {
            printf("%d",res[i]);
        }
        printf("\n");
    }
}