Simple simulation - Squirrel eat fruit

· This is a sample shows the wrong title; modify the sample description: 3,8,5,9.

· This is a sample of explanation, I hurt for half an hour.

Lowe valley entrance: P2069.

·topic:

Th one of N squirrels like to eat fruit sequence from the bottom up in a row, and reference numerals 1,2, ... N. A squirrel began to jump upwardly from the lowermost fruit, and the i-th hop skip can once i * i * i + 1 is divided by the number of fruit I 5 (= i * i * i% 5 + 1), and the foot eat fruit, fruit above if, under gravity, will fall down one space. The 1st hop skip from the first fruit 1 * 1 * 1 + 1% 5 = 2 fruit, skip to the third fruit, and the third eat fruit; from 2nd on 4 fruit (fruit falls third original position) skip 2 * 2 * 2 + 1% 5 = 4 to 8 of the fruit, and the eighth ate; so ....

Of course, there is always a squirrel jump out of the top of the string of fruit, to every K times, it can not eat any of the fruit. Then she returns to the bottom of the fruit, redo its K-th jump, in order to eat the fruit. So, ask it to eat only fruit of m (that is, the fruit of the M-hop eat) What label is?

· Ideas:

Array continues to fall.

The picture shows the dance through the first, second, third value of the number of groups.

· Code:

#include<iostream>
#include<cstring>
using namespace std;
int n,m,t;
int a[10005];
int yx(int a)
{
	return (a*a*a)%5+1;
}
void down(int x,int end)
{
	for(int i=x;i<=end;i++)
	{
		a[i]=a[i+1];
		}
}
int main()
{
	cin>>n>>m;
	int last;
	int step=1;
	t=n;
	for(int i=1;i<=n;i++)
	a[i]=i; 
	for(int i=1;i<=m;i++)
	{
		int s;
		s=yx(i);//跳过个数
		if(step+s<=t)//若没有超出界限
		{
		    last=step+s;//所吃果子的位置
		    step+=s;
		    if(i!=m)
			{t -; // coordinate position of the highest point - due to a fall, so a number of the highest point also fall 
		    down (step, t); // array fall 
				} 
			} 
		the else // If it exceeds the limit 
		{ 
			STEP =. 1; / / Skip lowermost first fruit 
		    Last + = STEP S; 
		    STEP + = S; 
		    IF (! I = m) 
			{ 
				T--; 
		    Down (STEP, T); 
				} 
			} 
		} 
	COUT << a [Last] << endl; 
	return 0; 
}