1058 A+B in Hogwarts (20)

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, 10^7^], 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

题目大意：就是一个进制转换的问题，最低位的进制是29，次低位的进制是17； 需要注意的是计算的时候，最高位溢出，开始用int来计算，始终有一个测试点事无法通过，修改为long long即可

 1 #include<iostream>
 2 #include<string>
 3 using namespace std;
 4 int main(){
 5   string a, b;
 6   cin>>a>>b;
 7   long long i,cnt=0, m[]={0,0,0}, n[]={0,0,0};
 8   for(i=0; i<a.size(); i++){
 9     if(a[i]!='.') m[cnt] = m[cnt]*10 + (a[i]-'0');
10     else cnt++;
11   }
12   cnt=0;
13   for(i=0; i<b.size(); i++){
14     if(b[i]!='.') n[cnt] = n[cnt]*10 + (b[i]-'0');
15     else cnt++;
16   }
17   long long numa=m[0]*29*17+m[1]*29+m[2]+n[0]*29*17+n[1]*29+n[2];
18   long long numb = numa%(29*17);
19   cout<<numa/29/17<<"."<<numb/29<<"."<<numb%29<<endl;
20   return 0;
21 }