Integer Inquiry
Time Limit: 2000/1000 MS (Java/Others)??? Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12499??? Accepted Submission(s): 3153
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
?
Source
East Central North America 1996
/*
题意：
将给出的大数相加起来，看数据即懂
*/
/*
100个100位的数相加，和不会超过102位的。如100个最大两位数99相加，也不过9900，四位数搞定。因此102个数组元素足够
为了减小内存使用动态数组；
*/
#include<cstdio>
#include<string.h>
#include<malloc.h>
#include<stdlib.h>
#define M 102
void add(char *a,int *sum)
{
?int i,j,l,alen;
?alen=strlen(a);
?
?int *num=(int*)malloc(sizeof(int)*alen);
?for(i=0;i<alen;i++)
?? num[i]=0;
??
?for(i=alen-1,j=0;i>=0;i--,j++)
????? num[j]=a[i]-'0';
?for(i=0;i<alen;i++)
? {
? ?sum[i]+=num[i];
? ?? if(sum[i]>9)
? ? {
? ?? sum[i]-=10;
???? sum[i+1]+=1;?
? ? }
? }
? free(num);???
}
int main() { ?int ca; ?scanf("%d",&ca); ?while(ca--) ?{ ?? int i=0; ?? char *a=(char*)malloc(sizeof(char)*M); ?? int? *sum=(int*)malloc(sizeof(int)*M); ?? for(i=0;i<M;i++) ????? sum[i]=0; ????? ?? while(scanf("%s",a)&&a[0]!='0') ???add(a,sum); ??? ???? ??for(i=M-1;i>=0&&sum[i]==0;i--) ?? ???????? ; ???????? if(i>=0) ???? for(;i>=0;i--) ????? printf("%d",sum[i]); ????? else ?????? printf("0"); ??? printf("n");? ???? ?if(ca!=0) ???? printf("n");? ???? free(a); ???? free(sum); ? ?} ?return 0; }
