ＨＤＵ A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 307 Accepted Submission(s): 147

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 字符串模拟大数加法 #include <iostream> #include <string.h> #include <stdio.h> using namespace std; char sum[1000]; void  add(string a,string b) {      string c ;     if(a.length()<b.length())     {c = a ; a = b; b = c;}     int l1 = a.length()-1;     int l2 = b.length()-1;     int k=0,i=0;     memset(sum,0,sizeof(sum));     char t;         while(l2!=-1)         {            t = (a[l1]-'0')+(b[l2]-'0');            sum[i]= (t+k)%10+'0';            k = (t+k)/10;            l1--;            l2--;            i++;         }          while(l1!=-1)          {              t = (a[l1]-'0');              sum[i] = (k+t)%10+'0';              k = (t+k)/10;              l1--;              i++;          }          if(k!=0)sum[i]=k+'0'; } int main() {     string a,b;     int T;     cin>>T;     for(int i = 1 ; i <= T ;i ++)     {         cin>>a>>b;         add(a,b);         cout<<"Case "<<i<<":"<<endl;         cout<<a<<" + "<<b<<" = ";         for(int j = strlen(sum)-1; j >=0;j--)         {             cout<<sum[j];         }         if(i<T)         cout<<endl<<endl;         else cout<<endl;     }     return 0; }