POJ 1879 Tempus et mobilius time and motion

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The main idea:

There is a timer, consisting of a ball numbered from 1 to N, with three tracks, corresponding to 1 minutes, 5 minutes, and 1 hours respectively. These three tracks can be thought of as stacks, each minute out of the heap of a ball, a ball heap can be considered a queue, the capacity of three tracks is 4,11,11. The ball first enters the 1-minute orbit, and when the fifth ball enters, the ball of the one-minute track is all out of the stack and into the team row. While the fifth ball enters the 5-minute orbit, the 5-minute track fills the ball into the hour track, and the 5-minute track ball all goes out of the stack into the team row. The hour is full and so is the track. At this time after half a day. Q: How many celestial spheres are returned to the original state.

Problem Solving Ideas:

Every half-day ball completes a sequence exchange, but asks how many days it takes to simulate the sequence exchange of a day's ball and then the least common multiple of the loop. (I can only simulate the sequence exchange of the ball in a day, but how to find out how many days it takes to get back to the original state or not to understand). For details, see the code:

#include <stdio.h> #include <string.h> #include <stack> #include <queue>using namespace Std;int NEXT[7005],MARK[7005]; Long Long gcd (long long A,long long B)//greatest common divisor {if (!b) return A;elsereturn gcd (b, a% b);} Long long LCM (long long X,long long y)//least common multiple {return x/gcd (x, y) * y;} int main () {int n;while (scanf ("%d", &n), n) {queue <int> q;stack <int> min,five,hour;for (int i=1;i<=n; i++) Q.push (i);//mark the ball into the queue for (int t=0;t<12*60*2;t++)//simulate the movement of the ball in the day {int now=q.front ();//each time it comes out is the ball of the first team Q.pop (); if ( Min.size () < 4)//If the 1 minute track is not full, enter this track Min.push (now), else{for (int i=1;i<=4;i++)//If this track is full, the ball into the team column {Q.push (Min.top ( )); Min.pop ();} if (Five.size () < 11)//Then determine whether the 5 minute ball is full, not full, then enter this track Five.push (now), else{for (int i=1;i<=11;i++)//If 5 minutes the ball is full, this track ball into the team column { Q.push (Five.top ()); Five.pop ();} if (Hour.size () < 11)//To determine the hour track Hour.push (now), else{for (int i=1;i<=11;i++) {Q.push (Hour.top ()); Hour.pop ();} Q.push (now);}}} for (int i=1;i<=n;i++)//Save the position of the ball in the day to the next array {next[i]=q.front (); Q.pop ();} MemseT (mark,0,sizeof (Mark)); Long long sum=1;for (int i=1;i<=n;i++)//Find minimum number of days to complete a loop {if (!mark[i]) {Long Long cnt=1;mark[i ]=1;int t=next[i];//The marking of the first ball to T while (!mark[t]) {mark[t]=1;cnt++;t=next[t];} SUM=LCM (sum,cnt);}} printf ("%d balls cycle after%lld days.\n", n,sum);}  return 0;}

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POJ 1879 Tempus et mobilius time and motion