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 }