A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 209179    Accepted Submission(s): 40226

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

就是一简单的大数，可是因为后面的换行符没有考虑周全，PE了六回

代码如下：

#include<stdio.h>
#include<string.h>
int main()
{
 int n,i,j,t,p,q,d,k;
 char a[1010],b[1010];
 int c[1010];
 scanf("%d",&n);
 //getchar();
 for(t=1;t<=n;t++)
 {
  scanf("%s%s",a,b);
  p=strlen(a);
  q=strlen(b);
  d=0;
  for(i=p-1,j=q-1,k=0;i>=0&&j>=0;i--,j--,k++)
  {
   d+=a[i]-'0'+b[j]-'0';
   c[k]=d%10;
   d/=10;
  }
  if(k==p)
  {
   while(j>=0)
   {
    d+=b[j]-'0';
    c[k]=d%10;
    d/=10;
    j--;k++;
   }
  }
  else
  {
   while(i>=0)
   {
    d+=a[i]-'0';
    c[k]=d%10;
    d/=10;
    i--;k++;
   }
  }
  if(d!=0)
  {
   c[k]=d;
   k++;
  }
  printf("Case %d:\n",t);
  printf("%s + %s = ",a,b);
  for(i=k-1;i>=0;i--)
  printf("%d",c[i]);
  printf("\n");
  if(t!=n)//此处就是控制的条件
  printf("\n");
 }
 return 0;
}

杭电 1002 A + B Problem II（大数处理）