A plus B II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 231970    Accepted Submission(s): 44505

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

2
1 2
112233445566778899 998877665544332211

 

Sample Output

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

 

参考代码

#include <stdio.h>#include <string.h>int main(){char str1[1001],str2[1001];int t, a[1001], b[1001],c[1001];int num,i,j,index1,index2,index,incre;scanf("%d",&t);for(num=1;num<=t;num++){scanf("%s",str1);scanf("%s",str2);for(i=0;i<1001;i++)a[i]=b[i]=0;for(i=0;i<strlen(str1);i++)a[i]=str1[strlen(str1)-1-i]-'0';for(i=0;i<strlen(str2);i++)b[i]=str2[strlen(str2)-1-i]-'0';index1=index2=index=0;incre=0;while(index1<strlen(str1) || index2<strlen(str2)){c[index]=(a[index1]+b[index2]+incre)%10;incre=(a[index1]+b[index2]+incre)/10;index1++;index2++; index++;}printf("Case %d:\n",num);printf("%s + %s = ",str1,str2);for(i=index-1;i>=0;i--)printf("%d",c[i]);printf("\n");}return 0;}