1058 A+B in Hogwarts (20 分)

1058 A+B in Hogwarts (20 分)

 

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], 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


加一下就可以。

#include <bits/stdc++.h>

using namespace std;
string s,ss;
int a[4],b[4],c[4];
int main(){
    cin >> s >> ss;
    s += '.';
    ss += '.';
    // cout << s << " " << ss << endl;
    int an = 0;
    int pos = 0;
    for(int i = 0; i < s.length(); i++){
        if(s[i] != '.'){
            an = an*10 + s[i] - '0';
        }else{
            a[pos++] = an;
            an = 0;
        }
    }
    pos = 0;
    an = 0;
    for(int i = 0; i < ss.length(); i++){
        if(ss[i] != '.'){
            an = an*10 + ss[i] - '0';
        }else{
            b[pos++] = an;
            an = 0;
        }
    }
    // cout << a[0]<<" "<<a[1]<<" "<<a[2]<<endl;
    // cout << b[0]<<" "<<b[1]<<" "<<b[2]<<endl;
    int flag = 0, flag1 = 0;
    for(int i = 2; i >= 0; i--){
        if(i == 2){
            c[i] = (a[i] + b[i])%29;
            flag = (a[i] + b[i])/29;
        }else if(i == 1){
            c[i] = (a[i] + b[i] + flag)%17;
            flag1 = (a[i] + b[i] + flag)/17;
        }else{
            c[i] = (a[i] + b[i] + flag1)%100000001;
        }
    }
    for(int i=0; i < 3; i++){
        printf("%d%c", c[i], i == 2?'\n':'.' );
    }

    return 0;
}