hd1002 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): 314201    Accepted Submission(s): 60894

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

 

#include<stdio.h>#include<string.h>#include<stdlib.h>int main(){char s1[1010];char s2[1010];int a[1010];int b[1010];int n,k=1;scanf("%d",&n);for(int l=1;l<=n;l++){int t,i,j;        scanf("%s%s",s1,s2);        printf("Case %d:\n",l);        printf("%s + %s = ",s1,s2);memset(a,0,sizeof(a));memset(b,0,sizeof(b));int len1=strlen(s1);int len2=strlen(s2);for(i=len1-1,j=0;i>=0;i--,j++)   a[j]=s1[i]-'0';for(i=len2-1,j=0;i>=0;i--,j++)   b[j]=s2[i]-'0';int len=len1>len2?len1:len2;for(i=0;i<len;i++){a[i]+=b[i];if(a[i]>9){a[i]-=10;a[i+1]++;}   }   while(len>0&&a[len]==0) len--;if(len==0)   printf("%d",a[0]);else   for(j=len;j>=0;j--)  printf("%d",a[j]);if(l!=n)   printf("\n\n");else   printf("\n");}return 0;}