hdu 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): 169696    Accepted Submission(s): 32544

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

 

#include<stdio.h>#include<string.h>int main(){char  str1[1000],str2[1000];int sum[1010],len1,len2,len;int i,k,n,c;scanf("%d",&n);for (c=1;c<=n;c++){memset(sum,0,sizeof(sum));memset(str1,0,sizeof(str1));memset(str2,0,sizeof(str2));len1=len2=len=0;scanf("%s%s",str1,str2);len1=strlen(str1);len2=strlen(str2);for(i=0,k=len1-1;i<len1;i++,k--)sum[k]=str1[i]-'0';for(i=len2-1,k=0;i>=0;i--,k++)sum[k]+=str2[i]-'0';if(len1>=len2)  len=len1;else  len=len2;for(i=0;i<len;i++){sum[i+1]+=sum[i]/10;sum[i]=sum[i]%10;}if(sum[len]>0)   len++;printf("Case %d:\n",c);printf("%s + %s = ",str1,str2);for (i=len-1;i>=0;i--)printf("%d",sum[i]);if(c<n)  printf("\n\n");else printf("\n");}return 0;}