codeforces 868B Race Against Time
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http://codeforces.com/problemset/problem/868/B
给出一个时间,对应时钟的时分秒针,再给出2个整点数,假设有2个杆子指向这2个整点数,问时分秒三针是否同时在这2个杆子的同一侧。
主要思路就是把时分秒针的位置用角度来表示,再遍历2个杆子的某一侧是否同时存在3根针或者是一根都没有,这2种情况为符合的情况。
#include<bits/stdc++.h>using namespace std;int main(){int h,m,s,t1,t2;while(cin>>h>>m>>s>>t1>>t2){ double ss=s*1.0*6; double mm=m*1.0*6+ss/360*6; double hh=h*30+mm/360*30; int t=t1*30; int tt=t2*30; int flag=0,flag1=0; if(ss==t||ss==tt)flag++; if(mm==t||mm==tt)flag++; if(hh==t||hh==tt)flag++; int sm=min(t,tt); int big=max(t,tt); if(ss>sm&&ss<big)flag1++; if(hh>sm&&hh<big)flag1++; if(mm>sm&&mm<big)flag1++; if(flag1==0||flag1+flag==3) cout<<"YES"<<endl; else cout<<"NO"<<endl;}return 0;}
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