Codeforces Round #438 B. Race Against Time 思路很重要

Have you ever tried to explain to the coordinator, why it is eight hours to the contest and not a single problem has been prepared yet? Misha had. And this time he has a really strong excuse: he faced a space-time paradox! Space and time replaced each other.

The entire universe turned into an enormous clock face with three hands — hour, minute, and second. Time froze, and clocks now show the time h hours, m minutes, s seconds.

Last time Misha talked with the coordinator at t1 o'clock, so now he stands on the number t1 on the clock face. The contest should be ready by t2 o'clock. In the terms of paradox it means that Misha has to go to number t2 somehow. Note that he doesn't have to move forward only: in these circumstances time has no direction.

Clock hands are very long, and Misha cannot get round them. He also cannot step over as it leads to the collapse of space-time. That is, if hour clock points 12 and Misha stands at 11 then he cannot move to 1 along the top arc. He has to follow all the way round the clock center (of course, if there are no other hands on his way).

Given the hands' positions, t1, and t2, find if Misha can prepare the contest on time (or should we say on space?). That is, find if he can move from t1 to t2 by the clock face.

Input

Five integers h, m, s, t1, t2 (1 ≤ h ≤ 12, 0 ≤ m, s ≤ 59, 1 ≤ t1, t2 ≤ 12, t1 ≠ t2).

Misha's position and the target time do not coincide with the position of any hand.

Output

Print "YES" (quotes for clarity), if Misha can prepare the contest on time, and "NO" otherwise.

You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").

某位大神的代码，基本思路和我一样，但是判断方法我真的麻烦到爆炸，最后错了一组数据.哎

这里是统计t1到t2通过了几个指针。  如果通过0个或者通过3个都是可以的。因为如果通过三个的话，反向就是0个。

一定不要for按个枚举..各种的错误..

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
#include <vector>
using namespace std;
int a,b,c,d,e;
int f[100];
int main()
{
    scanf("%d%d%d%d%d",&a,&b,&c,&d,&e);
    a*=5,d*=5,e*=5;
    a%=60;
    d%=60;
    e%=60;
    f[a]++;
    f[b]++;
    f[c]++;
    if(d>e) swap(d,e);
    int ans=0;
    for(int i=d;i<e;i++)ans+=f[i];
    if(ans%3==0)puts("YES");
    else puts("NO");
return 0;
}