1058. A+B in Hogwarts (20)

1058. A+B in Hogwarts (20)

时间限制

50 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

If you are a fan of Harry Potter, you would know the world of magic has its own currency system -- as Hagrid explained it to Harry, "Seventeen silver Sickles to a Galleon and twenty-nine Knuts to a Sickle, it's easy enough." Your job is to write a program to compute A+B where A and B are given in the standard form of "Galleon.Sickle.Knut" (Galleon is an integer in [0, 10⁷], Sickle is an integer in [0, 17), and Knut is an integer in [0, 29)).

Input Specification:

Each input file contains one test case which occupies a line with A and B in the standard form, separated by one space.

Output Specification:

For each test case you should output the sum of A and B in one line, with the same format as the input.

Sample Input:

3.2.1 10.16.27

Sample Output:

14.1.28

当相加为零时要处理。简单的另一种方法见

评测结果

测试点

#include<iostream>
#include<string>
using namespace std;  
long int Getnum(string s1,long int *l)
{ 
 long int temp=0,e=1;
 while((*l)>=0&&s1[*l]!='.')
 {
       temp=temp+(s1[*l]-'0')*e;
    e*=10; 
    (*l)--;
 }
 (*l)--; 
 return temp;
}
string Getstr(long int nn)
{
 string ss;
 char cc;  
 if(nn!=0)
 {
   while(nn>0)
   { 
  cc=nn%10+'0';
        ss=cc+ss;
  nn=nn/10; 
   } 
 }
 else
 {
  cc='0';
  ss=cc;
 }
 return ss;
}
int main()
{
  string A,B,sa;
  long int len1,len2,temp,c;
  cin>>A>>B;
  len1=A.length()-1;
  len2=B.length()-1; 
  c=0;
  temp=Getnum(A,&len1)+Getnum(B,&len2); 
  c=temp/29;
  temp%=29;
  sa="."+Getstr(temp);

  temp=c+Getnum(A,&len1)+Getnum(B,&len2);
  c=temp/17; 
  temp%=17;   
  sa="."+Getstr(temp)+sa;
 
  temp=c+Getnum(A,&len1)+Getnum(B,&len2);  
  sa=Getstr(temp)+sa;
  cout<<sa<<endl;
  return 0;
}