A + B Problem II hdoj 1002

A + B Problem II


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 209837    Accepted Submission(s): 40373




Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
 


Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
 


Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
 


Sample Input
2
1 2
112233445566778899 998877665544332211
 


Sample Output
Case 1:
1 + 2 = 3


Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
 


Author
Ignatius.L
 

一样的题目，不一样的做法；

#include<stdio.h>

#include<string.h>
#include<stdlib.h>
#define M 1000
char a[M];
char b[M];
char sum1[M];
char sum2[M];
int sum[M+100];

int main()
{
int T,ca;
scanf("%d",&T);
for(ca=1;ca<=T;ca++)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(sum1,0,sizeof(sum1));
memset(sum2,0,sizeof(sum2));
memset(sum,0,sizeof(sum));


int i,j,l,d,t;
scanf("%s%s",a,b);

for(i=strlen(a)-1,j=0;i>=0;i--,j++)
     sum1[j]=a[i]-'0';
     
    for(i=strlen(b)-1,j=0;i>=0;i--,j++)
      sum2[j]=b[i]-'0';
      
     l=strlen(a)>=strlen(b)?strlen(a):strlen(b); 
   
           for(i=0;i<l;i++)
             sum[i]=sum1[i]+sum2[i];
             for(i=0;i<l;i++)
             {
              if(sum[l-1]>9)
                l++;
                if(sum[i]>9)
                {
                sum[i]-=10;
                sum[i+1]+=1;
                }
             }
             
  printf("Case %d:\n",ca);
  printf("%s + %s  = ",a,b);

  for(i=l-1;i>=0;i--) 
        {
         printf("%d",sum[i]);
}
printf("\n");
   if(ca<T)
     printf("\n");
}
}