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;
}