poj1503 Integer Inquiry

最新推荐文章于 2019-03-23 17:24:53 发布

原创最新推荐文章于 2019-03-23 17:24:53 发布 · 202 阅读

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本文介绍了一种使用C++实现的大数加法算法，通过数组存储多位数字，并提供了示例代码来解决输入多个大整数并求和的问题。

题目描述：

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：大数加法!!

代码：

#include<iostream>
#include<string>
using namespace std;
int a[105],b[105];
char str[105];
int main()
{
    int i,j,len1=0,len2=0,temp1,temp2;
    memset(a,0,sizeof(a));
    while(gets(str))
    {
      if(str[0]=='0'&&strlen(str)==1){break;}
      len1=strlen(str);
      memset(b,0,sizeof(b));
      for(i=0;i<len1;i++) b[i]=str[i]-'0';
      if(len2==0)
      {
         for(i=0;i<len1;i++) {a[i]=b[i];}
         len2=len1;
      }
      else
      {
          if(len1>len2)
          {
            j=len2-1;
            for(i=len1-1;i>=0;i--)
            {
              if(j>=0){a[i]=a[j--]+b[i];}
              else a[i]=b[i];
            }
            len2=len1;
          }
          else
          {
            j=len1-1;
            for(i=len2-1;i>=0;i--)
            {
              if(j>=0){a[i]=a[i]+b[j--];}
              else break;
            }
          }
      }
    }
    temp1=0; temp2=0;
    for(i=len2-1;i>0;i--)
    {
       if(a[i]>=10) {a[i-1]+=a[i]/10;a[i]=a[i]%10;}
    }
    if(a[0]>=10){temp1=a[i]/100;temp2=(a[0]%100)/10;a[0]=a[0]%10;}
    if(temp1>0){printf("%d",temp1);}
    if(temp1>0&&temp2>=0){printf("%d",temp2);}
    for(i=0;i<len2;i++) {printf("%d",a[i]);}
    printf(" \n");
   // system("pause");
    return 0;
}