I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

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.

 

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.

 

 

 

Ignatius.L

#include<stdio.h>

#include<string.h>

 int toInt(char c){

     return c-'';

 }

 int main(){

     int i, n,j=,al,bl,ml,t;

     char a[], b[];//存储输入的两个数

    int count[];

     scanf("%d",&n);

     while(n--){

         if(j!=)printf("\n");

         scanf("%s",a);

         al=strlen(a);//返回字符串结束符之前的字符个数

         scanf("%s",b);

         bl=strlen(b);

         ml=(al>bl)?al:bl;//获取较长的字符长度

         t=ml;

         for(i=;i<=ml;i++)count[i]=;//初始化数组

        for(ml;al>&&bl>;ml--){

            count[ml]+=toInt(a[--al])+toInt(b[--bl]);

             if(count[ml]/){

                 //处理进位问题

                 count[ml-]++;

                 count[ml]=count[ml]%;

             }

        }

        while(al>){

            count[ml--]+=toInt(a[--al]);

            if(count[ml+]/){

                count[ml]++;

                count[ml+]%=;

            }

        }

         while(bl>){

             count[ml--]+=toInt(b[--bl]);

             if(count[ml+]/){

                 count[ml]++;

                 count[ml+]%=;

             }

         }

         printf("Case %d:\n%s + %s =",j++,a,b);

         for(i=;i<=t;i++){

             if(i==&&count[i]==){

                 i++;

             }

             printf("%d",count[i]);

         }

         printf("\n");

     }

     return ;

 }