A+B(Big Number Version)

Time Limit: 2000/1000ms (Java/Others)

Problem Description:

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 400.

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:

31 2112233445566778899 99887766554433221133333333333333333333333333 100000000000000000000

Sample Output:

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110Case 3:33333333333333333333333333 + 100000000000000000000 = 33333433333333333333333333 思路：400位数啊，明显是unusigned long long int是不够大的，所以就要自己写一个。所以我就想，把最后一位数相加，然后把十位数加到前一位数上（如果没有十位数，那就是0），最后把整个数输出来，因为不知道具体有多少位数，所以两个数的和我是倒着储存的（能看懂我的意思吗==） 难度：感觉有一定的难度，想了很长时间，主要是写的时候觉得有难度，想出来不算很难吧。要把字符串转变成整数数组。 代码：

1 #include
      
        2 #include
       
         3 int main() 4 { 5     int n; 6     while(scanf("%d",&n)!=EOF) 7     { 8         char ai[400],bi[400]; 9         int i,j,d,e,k,q=0,a[400],b[400],c[400];10         while(n--)11         {12             q++;13            getchar();14            scanf("%s",ai);15            scanf("%s",bi);16            d=strlen(ai);17            e=strlen(bi);18            for(i=0;i
        
         1||e>1)25            for(k=1,i=d-2,j=e-2;;i--,j--,k++)26            {27                if(i>=0&&j>=0)28                 c[k]=c[k-1]/10+a[i]+b[j];29               else if(i>=0&&j<0)30                c[k]=c[k-1]/10+a[i];31               else if(i<0&&j>=0)32                 c[k]=c[k-1]/10+b[j];33                else if(i<0&&j<0)34                {
    if(c[k-1]>=10)35                 c[k]=1;36                 else k--;break;}37            }38            printf("Case %d:\n",q);39            printf("%s + %s = ",ai,bi);40            for(;k>=0;k--)41            {c[k]=c[k]%10;42             printf("%d",c[k]);43             }44             printf("\n");45             if(n>0)46                 printf("\n");47         }48     }49     return 0;50 }