Integer Inquiry(大数相加）

最新推荐文章于 2016-10-11 21:31:00 发布

转载最新推荐文章于 2016-10-11 21:31:00 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/-lgh/p/4957629.html

本文介绍了一个用于处理和计算超长整数求和的C++程序设计方法。该程序能够接收最多100行输入，每行包含一个不超过100位的非常长的整数，并输出这些整数的总和。文章通过具体示例展示了如何实现这一功能。

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Problem 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

Sample Output

370370367037037036703703703670

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
char c[200],d[200],e[200];
void sum(char a[],char b[])
{
int al=strlen(a)-1;
int bl=strlen(b)-1;
int cl=0;
while(al>=0&&bl>=0)
{
c[cl]=c[cl]+a[al]+b[bl]-'0'*2;
c[cl+1]=c[cl+1]+c[cl]/10;
c[cl]=c[cl]%10;
al--;
bl--;
cl++;
}
while(al>=0)
{
c[cl]=c[cl]+a[al]-'0';
c[cl+1]=c[cl+1]+c[cl]/10;
c[cl]=c[cl]%10;
al--;
cl++;
}
while(bl>=0)
{
c[cl]=c[cl]+b[bl]-'0';
c[cl+1]=c[cl+1]+c[cl]/10;
c[cl]=c[cl]%10;
bl--;
cl++;
}
if(c[cl]==0)
cl--;
for(int i=cl,j=0; i>=0; i--,j++)
{
c[i]=c[i]+'0';
d[j]=c[i];
//printf("%c",c[i]);
}
//printf("\n");
//for(int i=0; i<=cl; i++)
//{
//printf("%c",c[cl-i]+'0');
//}printf("\n");
//return c[c1];
}
int main()
{
char a[150][150];
int i=0,j,k;
while(scanf("%s",a[i])!=EOF)
{
if(strcmp(a[i],"0")==0)
break;
i++;
}
strcpy(e,a[0]);
//if(strcmp(a[0],"0")==0)
//return 0;
for(j=1; j<i; j++)
{
memset(c,0,200);
sum(e,a[j]);
strcpy(e,d);
}
printf("%s\n",e);
return 0;
}