hdu acm-1047 Integer Inquiry(大数相加)

Integer Inquiry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11678    Accepted Submission(s): 2936

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

 

仅仅为了存一个高精度大数相加的模板

 #include <iostream>
 #include <stdio.h>
 #include <stdlib.h>
 #include <algorithm>
 #include <cstring>
 #include <math.h>
 #include <ctime>
 using namespace std;
 #define maxn 32768
 void add(char a[],char b[],char back[])
 {
     int i,j,k,up,x,y,z,l;
     char *c;
     if (strlen(a)>strlen(b)) l=strlen(a)+; else l=strlen(b)+;
     c=(char *) malloc(l*sizeof(char));
     i=strlen(a)-;
     j=strlen(b)-;
     k=;up=;
     while(i>=||j>=)
         {
             if(i<) x=''; else x=a[i];
             if(j<) y=''; else y=b[j];
             z=x-''+y-'';
             if(up) z+=;
             if(z>) {up=;z%=;} else up=;
             c[k++]=z+'';
             i--;j--;
         }
     if(up) c[k++]='';
     i=;
     c[k]='\0';
     for(k-=;k>=;k--)
         back[i++]=c[k];
     back[i]='\0';
 }

 int main()
 {
     int n;
     scanf("%d",&n);
     char a[maxn]="";
     while(n--)
     {
         char b[maxn],c[maxn];
         memset(b,'\0',sizeof(b));
         memset(c,'\0',sizeof(c));
         memset(a,'\0',sizeof(a));
         a[]='';
         while()
         {
             scanf("%s",b);
             if(b[]=='')
                 break;
             add(a,b,c);
             strcpy(a,c);
             memset(b,'\0',sizeof(b));
             memset(c,'\0',sizeof(c));
         }
         if(n==)
             printf("%s\n",a);
         else
             printf("%s\n\n",a);

     }
     return ;
 }