uva 424 Integer Inquiry(高精度加法)

本文介绍了一种使用高精度加法解决多个大整数求和的问题,通过自定义的数据结构实现对大整数的存储与运算,适用于整数长度不超过100位的情况。

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Integer Inquiry 

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 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.

Output

Your program should output the sum of the VeryLongIntegers given in the input.

Sample Input

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

370370367037037036703703703670

题目大意:多个数求和。

解题思路:高精度加法。

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

#define N 400

struct bign{
	int len, s[N];

	bign(){
		memset(s, 0, sizeof(s));
		len = 1;
	}

	bign operator = (const char *num){
		len = strlen(num);
		for (int i = 0; i < len; i++)
			s[i] = num[len - i - 1] - '0';
		return *this;
	}

	bign operator + (const bign &c){
		bign sum;
		sum.len = 0;

		for (int i = 0, g = 0; g || i < len || i < c.len; i++){
			int x = g;
			if (i < len) x += s[i];
			if (i < c.len) x += c.s[i];
			sum.s[sum.len++] = x % 10;
			g = x / 10;
		}

		return sum;
	}
	
};

int main(){
	char str[N];
	bign sum;
	while (cin >> str){
		if (strcmp(str, "0") == 0)
			break;

		bign num;

		num = str;

		sum = sum + num;

	}

	for (int i = sum.len - 1; i >= 0; i--)
		cout << sum.s[i];
	cout << endl;

	return 0;
}
		

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