PTA A+B in Hogwarts (20分)

It is the human mind that releases infinite light, and it is also the human mind that creates boundless darkness. Light and darkness are intertwined and fight together. This is the world for which we are nostalgic and helpless.

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

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#include<climits>//INT_MAX
#define PP pair<ll,int>
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3fll
#define dinf 1000000000000.0
#define PI 3.1415926
typedef long long ll;
using namespace std;
int const mod=1e9+7;
const int maxn=3e5+10;
ll S1[3],S2[3],fg[3],C[3]={10000000,17,29};
int main()
{
    scanf("%lld.%lld.%lld %lld.%lld.%lld", &S1[0], &S1[1], &S1[2], &S2[0], &S2[1], &S2[2]);
    for (ll i=2; i>=0; i--)
    {
        S1[i]+=S2[i]+fg[i];
        if (S1[i]>=C[i]&&i!=0)
        {
            fg[i-1]=S1[i]/C[i];
            S1[i]%=C[i];
        }
    }
    printf("%lld.%lld.%lld", S1[0], S1[1], S1[2]);
    return 0;
}