hdu1047 Integer Inquiry 多次大数相加

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1047

[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

纯大树相加：

代码例如以下：

#include <cstdio>
#include <cstring>
int sum[147];
char s[147];
void Add( char ss[])
{
int len = strlen(ss);
int z = 1;
for(int i = len - 1 ; i >= 0 ; i-- )
{
sum[z] +=(ss[i]-'0');
sum[z+1] +=sum[z]/10;
sum[z]%=10;
z++;
}
}

int main()
{
int t;
while(~scanf("%d",&t))
{
while(t--)
{
memset(sum,0,sizeof(sum));
while(scanf("%s",s)&&s[0]!='0')
{
Add(s);
}
int flag = 0;
for(int i = 147; i > 1 ; i-- )
{
if(flag == 0 && sum[i] == 0)
continue;
else
{
flag = 1;
printf("%d",sum[i]);
}
}
printf("%d\n",sum[1]);
if(t != 0)
printf("\n");
}
}
return 0;
}