HDU1047 Integer Inquiry

Integer Inquiry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1071    Accepted Submission(s): 240

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.
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.

题目大意：输入n组数据，每组数据中又有若干长度不大于100的整数，以0结束每组数据的输入，求每组中数据之和。

分析：

很明显，是大数问题，用字符数组接收输入的整数。然后模拟手工加法即可。

注意输出，由于输出出错郁闷了好几次。

代码：

#include<iostream>
#include<string>
using namespace std ;

int main (){

    char str [ 100 ];
    int x ,c ,n ,i ,j ,sum [ 110 ];
    cin >>n ;
    while (n --){
        memset (sum , 0 , sizeof (sum ));
        c = 0 ;
        while (cin >>str ,strcmp (str , "0" )){
            for (i =j =strlen (str )- 1 ; 0 <=i ;i --)
                sum [j -i ]+=str [i ]- '0' ;
        }
        for (i = 0 ;i < 110 ;i ++){
            x =sum [i ]+c ;
            sum [i ]=x % 10 ;
            c =x / 10 ;
        }
        i = 109 ;
        while (sum [i ]== 0 )i --;
        if (i < 0 )cout << "0" ;
        for (;i >= 0 ;i --)
            cout <<sum [i ];
        cout <<endl ;
        if (n )cout <<endl ;
    }
    return 0 ;
}