hdu 1047 Integer Inquiry 大数相加

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.

 

Sample Input

  
  

   
   1


123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0
  
  

 

Sample Output

  
  

   
   370370367037037036703703703670
  
  
  
  


  
  

题意： 一开始没看懂题目以为0代表整个程序输入结束，其实是标志了一个测试用例的结束。从这道题可以总结出一个规律：input部分描述的是一个测试用例的情况。如果题目中有This problem contains multiple test cases之类的话，表示有多个用例。

思路：核心的大数相加算法 : 每次读入一行数字到数组a中，然后从个位到高位将a的每个数字加到sum数组中。需要注意的是当a的每一位都加了后如果还有进位，要对每一位都进行处理，而不是只处理一次，因为可能会不断的向前进位。

代码：

#include <stdio.h>
#include <string.h>
#define N 105
char a[N],sum[N]; int  t,i,j,up,len; int main() {
    scanf("%d",&t); for(i=0; i<t; i++) {
        memset(sum,0,sizeof(sum)); do {
            scanf("%s",a); if(strcmp(a,"0")==0) {
                len=N-1; while(!sum[len]&&len>=0)len--; if(len==-1)printf("0"); for(j=len; j>=0; j--)printf("%d",sum[j]); break; }
            len=strlen(a);
            up=0; for(j=len-1; j>=0; j--) {
                sum[(len-1)-j] =up + sum[(len-1)-j] + (a[j]-'0') ;
                up=sum[(len-1)-j]/10;
                sum[(len-1)-j]%=10; } if(up) { for(; len-1-j<N; j--) {
                    sum[(len-1)-j]=up+sum[(len-1)-j];
                    up=sum[(len-1)-j]/10;
                    sum[(len-1)-j]%=10; } } } while(1);
        i!=t-1?printf("\n\n"):printf("\n"); } return 0; }