Integer Inquiry(大数相加)

最新推荐文章于 2016-10-11 21:31:00 发布

原创最新推荐文章于 2016-10-11 21:31:00 发布 · 523 阅读

0 ·

CC 4.0 BY-SA版权

大数处理专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一个使用超级计算机处理非常大的整数相加的问题。输入由一系列大整数组成，每个整数长度不超过100位，程序需正确处理这些大数并输出它们的总和。示例展示了如何通过逐位累加并处理进位来解决此类问题。

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

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
	char a[1000];	
	int i,j,x;
    while(cin>>a)
	{
		int sum[101]={0};
		while(1)
		{
			int b[100]={0};
			if(strcmp(a,"0")==0)
				break;
			int t=strlen(a);
			x=99;
			for(i=t-1;i>=0;i--)
				b[x--]=a[i]-'0';
			for(i=0;i<100;i++)
				sum[i]+=b[i];
			x=0;
			for(i=99;i>=0;i--)
			{
				sum[i]+=x;
				x=0;
			    if(sum[i]>9)
				{
					x=sum[i]/10;
				    sum[i]=sum[i]%10;
				}
			}
			cin>>a;
		}
		for(i=0;i<100;i++)
		{
			if(sum[i]!=0)
				break;
		}
		for(j=i;j<100;j++)
			cout<<sum[j];
		cout<<endl;
	}
	return 0;
}