Codeforces Round #438 (Div. 1 + Div. 2 combined) B. Race Against Time

B. Race Against Time

Problem statement

    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”).

Examples

Example 1
    Input
        12 30 45 3 11
    Output
        NO
Example 2
    Input
        12 0 1 12 1
    Output
        YES
Example 3
    Input
        3 47 0 4 9
    Output
        YES

Note

    The three examples are shown on the pictures below from left to right. The starting position of Misha is shown with green, the ending position is shown with pink. Note that the positions of the hands on the pictures are not exact, but are close to the exact and the answer is the same.

题意

    给您时针、分针和秒针的位置以及一个人在时钟上的起始点和终点，问你是否能不跨过任何一根针来从起点沿着圆盘走到终点

思路

    只要将时针，分针和秒针的位置都化为时、分、秒的样式，然后从起点到终点从两个方向一秒一秒遍历下去就行了。（做法有些奇特，至于如何转化详情请见代码）

Code

#include<bits/stdc++.h>using namespace std;typedef long long ll;inline void readInt(int &x) {    x=0;int f=1;char ch=getchar();    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();    x*=f;}inline void readLong(ll &x) {    x=0;int f=1;char ch=getchar();    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();    x*=f;}/*================Header Template==============*/int h[5],m[5],s[5];bool f1=1,f2=1;int main() {    readInt(h[2]);    readInt(h[3]);    readInt(h[4]);    readInt(h[0]);    readInt(h[1]);    m[0]=s[0]=m[1]=s[1]=0;    m[2]=h[3];    s[2]=h[4];    m[3]=(h[3]%5)*12;    int tmph3=0,tmph4=0;    if(m[3]>=60) {        tmph3+=m[3]/60;        m[3]%=60;    }//当时是这个if没加*1    s[3]=0;    m[4]=(h[4]%5)*12;    if(m[4]>=60) {        tmph4+=m[4]/60;        m[4]%=60;    }//当时是这个if没加*2    s[4]=0;    h[3]/=5;    h[3]+=tmph3;    if(h[3]==0)        h[3]=12;//当时是这个if没加*3    h[4]/=5;    h[4]+=tmph4;    if(h[4]==0)        h[4]=12;//当时是这个if没加*4//  for(int i=0;i<=4;i++)//      cout<<h[i]<<" "<<m[i]<<" "<<s[i]<<endl;    int x=h[0],y=m[0],z=s[0];    while(x!=h[1]||y!=m[1]||z!=s[1]) {//      cout<<x<<" "<<y<<" "<<z<<" "<<int(f1)<<endl;        for(int i=2;i<=4;i++) {            if(x==h[i]&&y==m[i]&&z==s[i]) {                f1=0;                break;            }        }        z++;        if(z==60) {            y++;            z=0;        }        if(y==60) {            x++;            y=0;        }        if(x==13)            x=1;    }    x=h[1],y=m[1],z=s[1];    while(x!=h[0]||y!=m[0]||z!=s[0]) {        for(int i=2;i<=4;i++) {            if(x==h[i]&&y==m[i]&&z==s[i]) {                f2=0;                break;            }        }        z++;        if(z==60) {            y++;            z=0;        }        if(y==60) {            x++;            y=0;        }        if(x==13)            x=1;    }    if(f1||f2)        puts("YES");    else        puts("NO");    return 0;}