杭电1002 A + B Problem II

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

 

//我以前也写过，这一次写出了新问题，可能你也会出现

#include<stdio.h>#include<string.h>#define max 1010int main(){int a[max],b[max];char s1[max],s2[max];int len1,len2,len,i,j;int t,k=0;scanf("%d",&t);getchar();while(t--){scanf("%s %s",s1,s2);printf("Case %d:\n",++k);printf("%s + %s = ",s1,s2);memset(a,0,sizeof(a));memset(b,0,sizeof(b));len1=strlen(s1);len2=strlen(s2);j=0;for(i=len1-1;i>=0;--i){a[j++]=s1[i]-'0';}j=0;for(i=len2-1;i>=0;--i){b[j++]=s2[i]-'0';}//len=len1>len2?len1:len2; for(i=0;i<max;++i)//要写成max ，不能写成len ，因为相加会进位 {a[i]+=b[i];if(a[i]>=10){a[i]-=10;a[i+1]++;}}for(i=max-1;i>0;--i)//这里i>0，0+0=0；不然没输出 {if(a[i]!=0) break;}for(j=i;j>=0;--j){printf("%d",a[j]);}puts("");if(t) puts("");}return 0;}