hdu 1002（大数）

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 318065    Accepted Submission(s): 61825

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

 

字符串数组模拟

 #include<iostream>
 #include<cstring>
 #include<string.h>
 #include<cmath>
 #include<algorithm>
 typedef long long ll;
 using namespace std;
 #define N 1010
 int main()
 {
     int n,k,i,j,flag;
     char s1[N],s2[N];
     int a[N],b[N];
     int c[N];
     cin>>n;
     for(k=;k<=n;k++)
     {
         cin>>s1>>s2;
         int len1=strlen(s1);
         int len2=strlen(s2);
         cout<<"Case "<<k<<":"<<endl<<s1<<" + "<<s2<<" = ";
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         int j=;
         int jj=;
         for(i=len1-;i>=;i--)
         {
             a[j]=s1[i]-'';
             j++;
         }
         for(i=len2-;i>=;i--)
         {
             b[jj]=s2[i]-'';
             jj++;
         }
         memset(c,,sizeof(c));
         for(i=;i<=max(len1-,len2-);i++)
         {
             c[i]=a[i]+b[i]+c[i];
             if(c[i]>)
             {
                 c[i]=c[i]-;
                 c[i+]++;
             }
         }
         flag=;
         for(i=max(len1-,len2-);i>=;i--)
         {
             if(flag==&&c[i]==)
                 flag=;
             else
             {
                 cout<<c[i];
                 flag=;
             }
         }
         if(k<n)
             cout<<endl<<endl;
         else
             cout<<endl;
     }
     return ;
 }