C++程序练习-Integer Inquiry-加法计算题

描述
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.)

输入
The input will consist of at most 100 lines of text, each of which contains a single VeryLongInteger. 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.

输出
Your program should output the sum of the VeryLongIntegers given in the input.
样例输入
123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0
样例输出
370370367037037036703703703670
提示
注意输入数据中，VeryLongInteger 可能有前导0

分析：
大数加法运算。此题为连续加法。

//冰非寒([e-mail])
//252kB 0ms 869 B G++ 
#include <iostream>
#include <string>
using namespace std;
class MyInteger{
public:
	MyInteger();
	void add(string);
	void print();
private:
	static const int N = 110;
	int res[N];
};
MyInteger::MyInteger(){
	for(int i(0);i < N;++ i){
		res[i] = 0;
	}
}
void MyInteger::add(string s){
	int temp[N]={0},j(0),slen = s.length();
	for(int i(slen - 1);i >= 0;-- i){
		temp[j ++] = s[i] - '0';
	}
	for(int i(0);i < N - 1;++ i){
		res[i] += temp[i];
		if(res[i] > 9){
			res[i + 1] += 1;
			res[i] %= 10;
		}
	}
}
void MyInteger::print(){
	bool output = false;
	for(int i(N - 1);i >= 0;-- i){
		if(res[i] > 0 || output){
			cout<<res[i];
			output = true;
		}
	}
	cout<<endl;
}
int main(){
	string s;
	MyInteger myint;
	while(cin>>s){
		if(0 == s.compare("0")){
			myint.print();
			break;
		}
		myint.add(s);
	}
	return 0;
}