HDU 1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 168135    Accepted Submission(s): 32205

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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

 

Author

Ignatius.L

代码：

#include<stdio.h>#include<string.h>char a[1000],b[1000];long c[1004];int main(){  long n,t,t1,t2,sign,i,k;   while(scanf("%ld%",&n)!=EOF)   {     t=0;      while(t<n)   {      scanf("%s%s",a,b);          t1=strlen(a)-1;       t2=strlen(b)-1;        for(k=0,sign=0;t1>=0&&t2>=0;t1--,t2--,k++)      if((a[t1]-'0')+(b[t2]-'0')+sign>=10)      {  c[k]=((a[t1]-'0')+(b[t2]-'0')+sign)%10;          sign=1;             }       else      {    c[k]=(a[t1]-'0')+(b[t2]-'0')+sign;           sign=0;       }                       if(t2+1==0)    { for(;t1>=0;t1--)    { if(a[t1]-'0'+sign>=10)      {  c[k++]=(a[t1]-'0'+sign)%10;         sign=1;             }       else      {    c[k++]=((a[t1]-'0')+sign)%10;           sign=0;       }          }    }    else if(t1+1==0)    {         for(;t2>=0;t2--)      if((b[t2]-'0'+sign)>=10)      {  c[k++]=(b[t2]-'0'+sign)%10;         sign=1;             }       else      {    c[k++]=((b[t2]-'0')+sign)%10;           sign=0;       }    }       if(sign) c[k++]=sign;    t++;          printf("Case %ld:\n%s + %s = ",t,a,b);    for(i=k-1;i>=0;i--)     printf("%ld",c[i]);    if(t==n)   printf("\n");              else printf("\n\n");        }          }     return 0;}