GDUFE ACM-1002

题目：http://acm.gdufe.edu.cn/Problem/read/id/1002

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:

3
1 2
112233445566778899 998877665544332211
33333333333333333333333333 100000000000000000000

Sample Output:

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

Case 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：

#include<stdio.h>
#include<string.h>
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        char ai[400],bi[400];
        int i,j,d,e,k,q=0,a[400],b[400],c[400];
        while(n--)
        {
            q++;
           getchar();
           scanf("%s",ai);
           scanf("%s",bi);
           d=strlen(ai);
           e=strlen(bi);
           for(i=0;i<d;i++)
            a[i]=ai[i]-'0';
           for(j=0;j<e;j++)
            b[j]=bi[j]-'0';
            k=0;
            c[0]=a[d-1]+b[e-1];
            if(d>1||e>1)
           for(k=1,i=d-2,j=e-2;;i--,j--,k++)
           {
               if(i>=0&&j>=0)
                c[k]=c[k-1]/10+a[i]+b[j];
              else if(i>=0&&j<0)
               c[k]=c[k-1]/10+a[i];
              else if(i<0&&j>=0)
                c[k]=c[k-1]/10+b[j];
               else if(i<0&&j<0)
               {if(c[k-1]>=10)
                c[k]=1;
                else k--;break;}
           }
           printf("Case %d:
",q);
           printf("%s + %s = ",ai,bi);
           for(;k>=0;k--)
           {c[k]=c[k]%10;
            printf("%d",c[k]);
            }
            printf("
");
            if(n>0)
                printf("
");
        }
    }
    return 0;
}