• 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 }
    有疑惑或者更好的解决方法的朋友,可以联系我,大家一起探讨。qq:1546431565
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  • 原文地址:https://www.cnblogs.com/mr-stn/p/9136772.html
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