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最新推荐文章于 2019-05-23 18:59:01 发布

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最新推荐文章于 2019-05-23 18:59:01 发布

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分类专栏：北大POJ 文章标签： random numbers integer c output input

本文链接：https://blog.youkuaiyun.com/xinghongduo/article/details/5875608

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本文探讨了一种简单的线性同余随机数生成器(RNG)，并提供了一个计算最长随机数序列间隙的C++实现。该算法通过不断迭代公式来生成随机数序列，并测量连续随机数之间的最大差距。

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Random Gap

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3538		Accepted: 871

Description

The pseudo-random number genegators (or RNGs for short) are widely used in statistical calculations. One of the simplest and ubiquitious is the linear congruence RNG, that calculates the n-th random number Rn as Rn = (aR_{n - 1} + c) mod m, where a, c and m are constants. For example, the sequence for a = 15, c = 7, m = 100 and starting value R0 = 1 is 1, 22, 37, 62, 37, 62, ...

Linear RNG is very fast, and can yield surprisinly good sequence of random numbers, given the right choice of constants. There are various measures of RNG quality, and your task is to calculate one of them, namely the longest gap between members of the sequence. More precisely, given the values of a, c, m, and R0, find such two elements Ri < Rj that:

there are no other Rk that Ri ≤ Rk ≤ Rj,
the difference Rj − Ri is maximal.

Input

Input contains integer numbers a c m R0.
0 ≤ a, c, R0 ≤ 10⁷, 1 ≤ m ≤ 16000000, am + c < 2³².

Output

Output should contain the single number -- the maximal difference found.

Sample Input

15 7 100 1

Sample Output

思路：这个题搞的头都大了。。。刚开始想用链表做，在有序链表中插入Rn，咋弄咋超时。。。最后不得不暴力了。。2***MS过了

看来暴力才是王道。。。::>_<::

#include<iostream>
using namespace std;
char lock[16000000];

int main()
{
	__int64 i,a,c,m,r0,temp;
	while(scanf("%I64d%I64d%I64d%I64d",&a,&c,&m,&r0)!=EOF)
	{
		memset(lock,0,sizeof(lock));
		temp=r0;
		while(!lock[temp])
		{
			lock[temp]=1;
			temp=(a*temp+c)%m;
		}
		int flag=0,max=0,j;
		for(i=0;i<16000000;i++)
		{
			if(lock[i]&&flag)
			{
				if(j>max)
					max=j;
				j=1;
			}
			if(lock[i]&&flag==0)
			{
				flag=1;	j=1;
			}
			if(lock[i]==0&&flag)
				j++;
		}
		cout<<max<<endl;
	}
	return 0;
}