题目：

有1,2,3,……无穷个格子，你从1号格子出发，每次1/2概率向前跳一格，1/2概率向前跳两格，走到格子编号为4的倍数时结束，结束时期望走的步数为____。

思路：

1、MonteCarlo模拟实验

参考代码

2、有限状态机的概率转移思想

跳一格跳两格都算一步； dp(i,j)表示从格子i到格子j的期望步数： dp(1,4)=1+0.5*dp(2,4)+0.5*dp(3,4)； dp(2,4)=1+0.5*dp(3,4)+0.5*dp(4,4); dp(3,4)=1+0.5*dp(4,4)+0.5*dp(1,4); dp(4,4)=0; 求解上述方程得到dp(1,4)=18/5;

代码：

#include<iostream>
#include<stdlib.h>
#include<time.h>
#include<string.h>

using namespace std;

bool rndJump(double p){
	double prob=rand()/(double)RAND_MAX;
	if(prob<=p)
		return true;
	else
		return false;
}

// MonteCarlo Simulation
double ExpectedSteps(){
	const int MAX_TRY=100000000;
	double expected=0;
	int total=0;
	int times=0;
	int steps=0;

	for(int i=0;i<MAX_TRY;i++){
		times=0;
		steps=1;
		while(steps%4!=0){
			if(rndJump(0.5))
				steps+=1;
			else
				steps+=2;
			times++;
		}
		total+=times;
	}

	expected=(double)total/MAX_TRY;

	return expected;
}


// DynamicProgramming-like
// f(1,4)=1+0.5*f(2,4)+0.5*f(3,4);
// f(2,4)=1+0.5*f(3,4)+0.5*f(4,4);
// f(3,4)=1+0.5*f(4,4)+0.5*f(1,4);
// f(4,4)=0;
double ExpectedSteps_DP(int n){
	double A[n],B[n];
	memset(A,0,n*sizeof(double));
	memset(B,0,n*sizeof(double));
	A[3]=0;
	B[3]=0;
	A[2]=1+0.5*A[3];
	B[2]=0.5;
	A[1]=1+0.5*A[2]+0.5*A[3];
	B[1]=0.5*B[2]+0.5*B[3];
	A[0]=1+0.5*A[1]+0.5*A[2];
	B[0]=0.5*B[1]+0.5*B[2];

	return (double)A[0]/(1-B[0]);
}


int main(){
	cout<< ExpectedSteps()<<endl;
	cout<< ExpectedSteps_DP(4)<<endl;
	return 0;
}