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最新推荐文章于 2019-04-29 21:06:00 发布

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最新推荐文章于 2019-04-29 21:06:00 发布

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本文链接：https://blog.youkuaiyun.com/weixin_44008424/article/details/86653786

本文探讨了使用超级计算机处理大规模计算任务的案例，特别关注于对VeryLongIntegers进行求和运算的算法实现。通过细致的代码解析，展示了如何处理长度可达100位的大整数相加问题。

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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. “Thissupercomputerisgreat,”remarkedChip. “IonlywishTimothywereheretoseetheseresults.” (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.

这题思路挺简单的，主要是细心



#include <iostream>
#include<cstring>
using namespace std;
int in[110][110];
int convert(int size)

{
    int maxn = 0;
    for (int i = 0; i < size; i++)
    {
         int len =strlen(in[i]);
         if (len > maxn)
             maxn= len;
         for (int j = 0; j < len /2; j++)
         {
             char temp = in[i][j];
             in[i][j]= in[i][len - 1 - j];
             in[i][len- 1 - j] = temp;
         }
    }
    return maxn;
}

 
int main()
{
    int line = 0;
    while (cin >>in[line] && in[line][0] != '0')
         line++;
    int carry[102] = { 0 };
    int size =convert(line);
    int result[110];
    for (int i = 0; i < size;i++)
    {
         int sum = 0;
         for (int j = 0; j < line;j++)
         {
             if (in[j][i] == '\0')
                  in[j][i]= '0';
             sum= sum + int(in[j][i]- '0');
         }
         sum= sum + carry[i];
         result[i]= sum % 10;
         carry[i+ 1] = sum / 10;
    }
    result[size]= carry[size];
    for (int j = 0; j < (size+ 1) / 2; j++)
    {
         int temp = result[j];
         result[j]= result[size - j];
         result[size- j] = temp;
    }
    for (int i = 0; i <=size; i++)
    {
         if (i == 0 &&result[i] == 0) continue;
         cout<< result[i];
    }
    cout<< endl;
}