1503. Integer Inquiry

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#水题 #大整数加法

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本文介绍了一个简单的程序设计问题——如何在一个超级计算机上处理并计算多个非常大的整数的总和。通过逆序读取每个大数的每一位，并进行逐位相加的方式处理进位，最终实现了大数求和的功能。

Description

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

思路

这道题比较简单，单独考虑每一位的相加，处理一下进位就ok了！
注意：

求和的sum数组大小至少sum[102];
题目中有no VeryLongInteger will be negative 这样的描述，所以输入中可能有以0开头的数，判断输入结束应该使用strcmp(a,”0”)语句，而非a[0]!=0;

代码

#include <iostream>
#include <string.h>

using namespace std;

int main()
{
    char a[101];
    int sum[102] = { 0 };
    int i = 0,j = 0;

    cin >> a;
    while (strcmp(a,"0"))
    {
        for (i = 0; i < strlen(a); i++)
        {
            sum[i] += a[strlen(a) - i - 1] - '0';
            if (sum[i] >= 10)   //进位的处理
            {
                sum[i] -= 10;
                sum[i + 1] += 1;
                j = i + 1;
            }
            else {
                j = i;
            }   
        }
        cin >> a;
    }

    for (i = j; i >= 0; i--)  //将sum逆序输出
    {
        cout << sum[i];
    }

    return 0;
}