标签：and seve UNC ecif lin mes pre ace 1.2
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)).
Each input file contains one test case which occupies a line with A and B in the standard form, separated by one space.
For each test case you should output the sum of A and B in one line, with the same format as the input.
3.2.1 10.16.27
14.1.28
long long输出是%lld
#include<iostream> #include<vector> #include<algorithm> #include<queue> #include<string> #include<map> #include<set> #include<stack> #include<string.h> #include<cstdio> #include<cmath> using namespace std; struct Node { long long Galleon; int Sickle,Knut; }; void add(Node&A,Node&B) { Node sum; int temp=0; sum.Knut=A.Knut+B.Knut; if(sum.Knut>=29) { sum.Knut-=29; temp=1; } sum.Sickle=A.Sickle+B.Sickle+temp; temp=0; while(sum.Sickle>=17) { sum.Sickle-=17; temp++; } sum.Galleon=A.Galleon+B.Galleon+temp; cout<<sum.Galleon<<"."<<sum.Sickle<<"."<<sum.Knut<<endl; } int main() { Node A[2]; char c; for(int i=0;i<2;i++) // cin>>A[i].Galleon>>c>>A[i].Sickle>>c>>A[i].Knut; scanf("%lld.%d.%d",&(A[i].Galleon),&(A[i].Sickle),&(A[i].Knut)); //cout<<A[0].Galleon<<" "<<A[0].Sickle<<" "<<A[0].Knut<<endl; add(A[0],A[1]); return 0; }
标签：and seve UNC ecif lin mes pre ace 1.2
原文地址：https://www.cnblogs.com/zhanghaijie/p/10324875.html
