高精度运算-加法

最新推荐文章于 2024-05-12 16:31:11 发布

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最新推荐文章于 2024-05-12 16:31:11 发布

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分类专栏： ACM比赛题目及总结

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本博客介绍了一个利用超级计算机处理VeryLongInteger求和问题的程序设计案例。Chip Diller使用BIT的新超级计算机探索了3的幂次方，并对这些数进行各种求和操作。文中提供了一个C++代码示例，展示了如何读取多行文本输入的VeryLongInteger并计算它们的总和。

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One of the first users of BIT’s new supercomputer was Chip Diller. He extended his exploration of
powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.
“This supercomputer is great,” remarked Chip. “I only wish Timothy were here to see these results.”
(Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky
apartments on Third Street.)
Input
The input will consist of at most 100 lines of text, each of which contains a single VeryLongInte-
ger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits (no
VeryLongInteger will be negative).
The final input line will contain a single zero on a line by itself.
Output
Your program should output the sum of the VeryLongIntegers given in the input.
Sample Input
123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0
Sample Output
370370367037037036703703703670

我的代码：

#include <iostream> 
#include <cstring>  
#include <string>  
using namespace std;
int main()
{
	string str;
	int a[250] = { 0 }, k, i, b[250] = { 0 };
	while (cin >> str)
	{
		if (str[0] == '0')break;
		a[0] = str.length();
		for (int i = 1; i <= str.size(); i++)
			a[i] = str[a[0] - i] - '0';
		for (int i = 1; i <= str.size(); i++)
		{
			int k = a[i] + b[i];
			b[i] = k % 10;
			b[i + 1] += k / 10;
		}
	}
	for (i = 250; i >= 0; i--)
		if (b[i])
			break;
	for (; i > 0; i--)
		cout << b[i];
	cout << endl;




}