1002 hdu (高精度)

A + B Problem II

TimeLimit: 2000/1000 MS(Java/Others)    MemoryLimit: 65536/32768 K (Java/Others)
Total Submission(s):108726    AcceptedSubmission(s): 20548

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 integerT(1<=T<=20) which means the number oftest cases. Then T lines follow, each line consists of two positiveintegers, A and B. Notice that the integers are very large, thatmeans you should not process them by using 32-bit integer. You mayassume 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 isthe an equation "A + B = Sum", Sum means the result of A + B. Notethere are some spaces int the equation. Output a blank line betweentwo test cases.

 

Sample Input

2 1 2112233445566778899 998877665544332211

 

Sample Output

Case1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 =1111111111111111110

 

Author

Ignatius.L

#include <stdio.h>#include<string.h> int sum[1000];void add (char a[] ,char b[]){     int i= strlen(a) - 1 ,j = strlen (b) - 1;     int x= 0 ,s ,k = 0;     while(i >= 0 && j>= 0)    {          s = a[i--]+ b[j--] - 96 + x ;          sum[k++] =s % 10 ;           x = s / 10;    }     if (j== -1)     while(i >= 0)    {         s = a[i--] - 48 + x ;         sum[k++] = s % 10 ;         x = s / 10 ;    }     if (i== -1)     while(j >= 0)    {         s = b[j--] - 48 + x ;         sum[k++] = s % 10 ;         x = s / 10 ;    }     if(x)     sum[k]= x ;    else      k = k- 1 ;          for (i= k ;i >= 0 ;i--)     printf("%d",sum[i]);}int main (){     int t,i;     chara[1000] ,b[1000];     scanf("%d",&t);     for (i= 1 ;i <= t ;i++)    {         memset(a , '0' ,sizeof(a));         memset(b , '0' ,sizeof(b));         scanf ("%s%s",a ,b);         printf ("Case %d:\n%s + %s =",i ,a ,b);         add (a ,b);         printf ("\n");         if(i != t)printf("\n");    }     return0;}