Integer Inquiry
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15325Accepted Submission(s): 3933
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
Source
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 105char a[N];int b[N],c[N];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]);elseprintf("0");printf("\n");if(n!=0)//两个输出结果之间要空一行！printf("\n");}return 0;}
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