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

 

 大数相加的问题，开两个数组分别储存a、b，再开一个数组sum用来储存最终的和值。

这题要注意相加时储存的顺序和最后输出时的顺序不要搞错。

当然，这里特别强调一下最后的输出格式，最后一组数据只要换一行！一行！行！原谅愚蠢如我在这个点被坑了5次！！还是刷题太少。。

 

附AC代码

1 #include
     
       2 #include
      
        3 const int MAX=1050; 4 int main() 5 { 6     int n,m,len1,len2,maxlen,minlen; 7     char a[MAX],b[MAX],sum[MAX];//三个主要数组  8     char *max, *min; 9     scanf("%d",&n);10     for(m=1;m<=n;m++)11     {12         int i,j,flag=0;13         memset(sum,0,sizeof(sum));14         scanf("%s %s",&a,&b);15         len1=strlen(a);16         len2=strlen(b);17         if(len1>=len2)//判断长度，长的为max 18         {19             maxlen=len1;20             minlen=len2;21             max = a;22             min = b;23         }24         else25         {26             maxlen=len2;27             minlen=len1;28             max = b;29             min = a;30         }31         for(i=maxlen-1,j=minlen-1;i>=maxlen-minlen&&j>=0;i--,j--)//长度重叠部分相加 32         {33             sum[i]+=max[i]+min[j]-'0'-'0';//转换为十进制整数，下面的同理 34             if(i==0&&sum[0]>9)//如果两数组长度相等 35             {36                 flag=1;//加在sum前表示进的1 37                 sum[i]-=10;//本大于9，进1故减10 38             }39             else if(sum[i]>9)40             {41                 sum[i-1]++;//进1 42                 sum[i]-=10;43             }44         }45         for(i=maxlen-minlen-1;i>=0;i--)//剩下的max加到sum里 46         {47             sum[i]+=max[i]-'0';48             if(i==0&&sum[0]>9)49             {50                 flag=1;51                 sum[i]-=10;52             }53             else if(sum[i]>9)54             {55                 sum[i-1]++;56                 sum[i]-=10;57             }58         }printf("Case %d:\n",m);59         60         if(flag)        61         {62             63             printf("%s + %s = 1",a,b);}64         else65         {66             67             printf("%s + %s = ",a,b);68         }69         for(i=0;i