大数专题A

#include <stdio.h>#include <string.h>main(){char a[1010],b[1010];int T,k,aa[1010],bb[1010],cc[1010],lena,lenb,lenc,i,x,flag=0;////freopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);scanf("%d",&T);for(k=1;k<=T;k++){memset(a,'\0',sizeof(a));memset(b,'\0',sizeof(b));scanf("%s%s",&a,&b);lena=strlen(a);lenb=strlen(b);memset(aa,0,sizeof(aa));memset(bb,0,sizeof(bb));memset(cc,0,sizeof(cc));for(i=0;i<lena;i++)aa[lena-i-1]=a[i]-48;//把lena-i-1改成lena-i就会报错for(i=0;i<lenb;i++)bb[lenb-i-1]=b[i]-48;lenc=0;x=0;while(lenc<lena||lenc<lenb){cc[lenc]=aa[lenc]+bb[lenc]+x;x=cc[lenc]/10;cc[lenc]%=10;lenc++;}cc[lenc]=x;if(cc[lenc]==0)lenc--;if(flag)printf("\nCase %d:\n",k);if(!flag){flag=1;printf("Case %d:\n",k);}printf("%s + %s = ",a,b);for(i=lenc;i>=0;i--)printf("%d",cc[i]);putchar('\n');}return 0;}

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