Integer Inquiry hdoj 1047

Integer Inquiry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15216    Accepted Submission(s): 3909

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

11234567890123456789012345678901234567890123456789012345678901234567890123456789012345678900

 

Sample Output

370370367037037036703703703670
/*题意：：     首先输入一个正整数，表示可以计算几组数，接下来是输入不大于100行的  字符串，直到输入的字符串的值为‘0’时，计算以上输入的字符串的和。  然后将其和输出。解题思路：：      这是个大数累加题，要用到大数加法。      首先定义一个字符数组（足够大，本题范围为0~100000）来存放连续输入的  字符串，然后用大数加法将各个位上的数累次相加。  但要注意输出格式：第一次输入与输出之间没有空格，其他的要加空格。 */#include<stdio.h>#include<string.h>#include<math.h>int a[100001];char s[100001];int main(){ int t,n,m,i,l; while(scanf("%d",&t)!=EOF) {  while(t--)  {   memset(a,0,sizeof(a));   while(scanf("%s",s)&&s[0]!='0')   {    n=1;    l=strlen(s);    for(i=l-1;i>=0;i--)//基本大数相加算法模型     {     a[n]+=s[i]-'0';     a[n+1]+=a[n]/10;     a[n]%=10;     n++;    }   }   m=0;   for(i=1001;i>1;i--)   {    if(a[i]==0&&m==0)        continue;    else    {     m=1;     printf("%d",a[i]);    }   }   printf("%d\n",a[1]);//输出格式    if(t)    printf("\n");  } } return 0;}