大数系列——Integer Inquiry

Integer Inquiry

One of the first users of BIT's new supercomputer was Chip Diller. Heextended his explorationof powers of 3 to go from 0 to 333 and he explored taking varIoUs sumsof those numbers.

``This supercomputer is great,'' remarked Chip. ``I only wish Timothy werehere to see theseresults.'' (Chip moved to a new apartment,once one became available onthe third floor of theLemon Sky apartments on Third Street.)

Input

The input will consist of at most 100 lines of text,each of whichcontains a single VeryLongInteger.Each VeryLongInteger will be 100 or fewer characters in length,and willonly 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

字符串系列刷得实在蛋疼，先换大数系列刷刷。。。

第一题很水，不愧是大数。

大数相加，倒序输入数组再逐位计算，再倒序输出。

AC代码：

#include<iostream>
#include<string>
using namespace std;

int main()
{
	int a[100] = {0},sum[110] = {0};
	string str;
	while (1)
	{
		cin >> str;
		if (str[0] == '0')
			break;
		for (int i = 0; i < (int)str.size(); i++)
			a[i] = str[(int)str.size() - i - 1] - '0';
		for (int i = 0; i < (int)str.size(); i++)
		{
			int temp = a[i] + sum[i];
			sum[i] = temp % 10;
			sum[i + 1] += temp / 10;
		}
	}
	int i;
	for (i = 100; i >= 0; i--)
		if (sum[i])
			break;
	for (;i >= 0; i--)
		cout << sum[i];
	cout << endl;
	return 0;
}