算法学习【4】 - 1381. a*b_2006 algorithm course examination-优快云博客

本文链接：https://blog.youkuaiyun.com/BetaBin/article/details/7356995

本文介绍了一种处理大整数乘法的方法，通过将大整数分解为字符串，并使用叠加的方式实现大整数与较小整数之间的乘法运算。此方法适用于a范围在0<=a<=10^100，b范围在0<=b<=10,000的情况。

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题目：http://soj.me/1381

1381. a*b

Description

Give two positive integers a and b, please help us calculate a*b.

Input

The first line of the input is a positive integer T. T is the number of test cases followed.

Each test case contain two integer a,b (0<=a<=10^100, 0<=b<=10,000) given in one line.

Output

The output of each test case should consist of one line, contain the result of a*b.

Sample Input

12 7

Sample Output

Problem Source

Algorithm Course Examination 2006

思路：大整数乘法，稍微修改了下，把b当成int叠加吧，感觉这种方法比较简单。

代码：

#include <iostream>
#include <string>

using namespace std;

string BigAdd(string a, string b)
{
	while (a.size() < b.size())
	{
		a = '0' + a;
	}
	while (a.size() > b.size())
	{
		b = '0' + b;
	}
	a = '0' + a;
	b = '0' + b;

	for (int i = a.size() - 1; i > 0; i--)
	{
		a[i] = a[i] + b[i] - '0';
		if (a[i] > '9')
		{
			a[i] = a[i] - 10;
			a[i - 1] = a[i - 1] + 1;
		}
	}

	int zeroNum = 0;
	for (int j = 0; j < a.size(); j++)
	{
		if (a[j] == '0')
		{
			zeroNum++;
		} 
		else
		{
			break;
		}
	}
	if (zeroNum == a.size())
	{
		a = '0';
	} 
	else
	{
		a = a.substr(zeroNum, a.size() - zeroNum);
	}

	return a;
}

string specialMul(string a, int b)
{
	string sum = "0";
	for (int i = 0; i < b; i++)
	{
		sum = BigAdd(sum, a);
	}
	return sum;
}

int main()
{
	int size;
	string a;
	int b;

	cin >> size;

	while (size--)
	{
		cin >> a >> b;
		cout << specialMul(a, b) << endl;
	}

	return 0;
}