HDU 1047 Integer Inquiry【大数】

Integer Inquiry

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 15325 Accepted Submission(s): 3933



[align=left]Problem Description[/align]
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.)

[align=left]Input[/align]
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.

[align=left]Output[/align]
Your program should output the sum of the VeryLongIntegers given in the input.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

[align=left]Sample Input[/align]

1

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

[align=left]Sample Output[/align]

370370367037037036703703703670

[align=left]Source[/align]
East Central North America 1996

思路：

大数多个数相加和一般的数多个数相加方法一样，都是用一个数来保存结果（数的和），将数如的数加到这个结果上直到不满足条件为止，输出结果！区别在于：大数需要用到数组，需要将大数的两个数的加法的算法用到这个上面，具体操作看代码！

代码：

#include <stdio.h>
#include <string.h>
#define N 105
char a
;
int b
,c
;
int main()
{
int n,i,j,k,len;
scanf("%d",&n);
while(n--)
{
memset(c,0,sizeof(c));//数组c用来保存最终的结果（多个大数的和）
while(gets(a)&&a[0]!='0')
{
len=strlen(a);
memset(b,0,sizeof(b));//数组b用来保存数组a中的字符-‘0’所得到的数值！
for(i=len-1,j=0;i>=0;i--)
{
b[j++]=a[i]-'0';
}
for(i=0;i<101;i++)//每次都将b加到c上
{
c[i]+=b[i];
if(c[i]>=10)
{
c[i]-=10;
c[i+1]++;
}
}
}
for(i=100;i>=0&&c[i]==0;i--);//输出c！
if(i>=0)
for(;i>=0;i--)
printf("%d",c[i]);
else
printf("0");
printf("\n");
if(n!=0)//两个输出结果之间要空一行！
printf("\n");
}
return 0;
}