hdu 1209 Clock

这个题不难但是里面有很多东西需要注意,尤其是Ties are broken in such a way that an earlier time precedes a later time.这就是说当夹角一样时,事件小的放在前面。

比较麻烦啊。

#include<iostream>
#include<string>
#include<math.h>
#include<algorithm>
using namespace std;
struct node{
	string ko;
	int shi;
	int fen;
	double angle;
};
bool cmp(node a,node b)
{
	if(a.angle!=b.angle)
		return a.angle<b.angle;
	else if(a.shi!=b.shi)
		return a.shi<b.shi;
	else return a.fen<b.fen;
}
node fuck[6];
int main()
{
	int T,i;
	string a;
	cin>>T;
	while(T!=0)
	{
		for(i=1;i<=5;i++)
		{
			cin>>a;
			fuck[i].ko=a;
			fuck[i].shi=(a[0]-'0')*10+(a[1]-'0');
			fuck[i].fen=(a[3]-'0')*10+(a[4]-'0');
		}
		for(i=1;i<=5;i++)
		{
			fuck[i].angle=abs(fuck[i].fen*6-(fuck[i].shi%12)*30-fuck[i].fen*0.5);
			if(fuck[i].angle>180)
				fuck[i].angle=360.0-fuck[i].angle;
		}
		sort(fuck+1,fuck+5+1,cmp);
		cout<<fuck[3].ko<<endl;
		T=T-1;
	}
	return 0;
}


 

原文地址:https://www.cnblogs.com/javawebsoa/p/2995495.html