A+B长字符串

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>#define N 1001char A[N],B[N],sum[N];int main(){int T,i,j,k,x,carry;    scanf("%d",&T);for(i=0;i<T;i++){if(i)printf("\n");scanf("%s%s",&A,&B);j=strlen(A)-1;k=strlen(B)-1;for(x=0,carry=0;(j+1)&&(k+1)!=0;j--,k--,x++){if((A[j]-'0')+(B[k]-'0')+carry<10){sum[x]=(A[j]-'0')+(B[k]-'0')+carry;carry=0;}else{sum[x]=(A[j]-'0')+(B[k]-'0')+carry-10;carry=1;}}if(j+1){for(;j>=0;j--,x++){if(A[j]-'0'+carry<10){sum[x]=(A[j]-'0')+carry;carry=0;}else{sum[x]=0;carry=1;}}}else if(k+1){for(;k>=0;k--,x++){if(B[k]-'0'+carry<10){sum[x]=(B[k]-'0')+carry;carry=0;}else{sum[x]=0;carry=1;}} }  if(carry) sum[x]=1; else x--; printf("Case %d:\n",i+1); printf("%s + %s = ",A,B); while(x>-1) printf("%d",sum[x--]); printf("\n");}return 0; }