121 另类乘法

最新推荐文章于 2021-07-04 22:43:22 发布

小可yaoyao

最新推荐文章于 2021-07-04 22:43:22 发布

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本文链接：https://blog.youkuaiyun.com/xiaokeyaoyao/article/details/9851823

本文介绍了一种独特的乘法算法，通过将两个整数的每一位进行相乘并求和来计算乘积。该方法利用字符串操作和字符转数字转换，简化了传统的乘法过程，适合对数学算法有兴趣的学习者和开发者。

另类乘法

时间限制：3000 ms | 内存限制：65535 KB

难度：1

描述

Bessie is tired of multiplying pairs of numbers the usual way, so she invented her own style of multiplication. In her style, A*B is equal to the sum of all possible pairwise products between the digits of A and B. For example, the product 123*45 is equal to 1*4 + 1*5 + 2*4 + 2*5 + 3*4 + 3*5 = 54. Given two integers A and B (1 ≤ A, B ≤ 1,000,000,000), determine A*B in Bessie's style of multiplication.

输入

The first Line of the input is a positive integer T,indicates the number of the test cases;
In every case,the input is in one line,contains two positive interger A,B

输出

For every case,output the multiplication in Bessie's style.

样例输入

1
123 45

样例输出

#include<stdio.h>
#include<string.h>
int main()
{
    int n,i,len1,len2,sum1,sum2,s[12],t[12];
    char a[12],b[12];
    while(scanf("%d",&n)!=EOF)
    {
        while(n--)
        {
            scanf("%s%s",a,b);
            len1=strlen(a);
            len2=strlen(b);
            sum1=sum2=0;
            for(i=0;i<len1;i++)
            {
                s[i]=a[i]-'0'; //将字符数字转化成数字
                sum1+=s[i];
            }
            for(i=0;i<len2;i++)
            {
                t[i]=b[i]-'0';
                sum2+=t[i];
            }
            printf("%d\n",sum1*sum2);
        }
    }
    return 0;
}