我试图解决Project Euler中的280th问题,为此我编写了以下模拟;
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <sys/time.h>
/* Directions
1
2 3
4
*/
int grid[5][5] = {
{0, 0, 0, 0, 2},
{0, 0, 0, 0, 2},
{0, 0, 0, 0, 2},
{0, 0, 0, 0, 2},
{0, 0, 0, 0, 2}
};
int InitPos[2] = {2, 2};
int MaxExp = 5000000;
bool Success = false;
int StepCount = 0;
int ExpNumber = 1;
int AntsBag = 0;
void Init();
void CarryFood(int * pos);
void LeftFood(int * pos);
bool checkMovability(int * pos, int direction);
bool moveToDirection(int pos[2], int direction);
bool checkSuccess();
void ShowResult();
int main(int argc, char const *argv[])
{
timeval curTime;
gettimeofday(&curTime, NULL);
int milli = curTime.tv_usec / 1000;
time_t t;
srand((unsigned)time(&t));
//timeTData*.txt corresponds to using "time(&t)" above
//milliData.txt corresponds to using "milli" variable above
//timeTUnsigData*.txt corresponds to using "(unsigned)time(&t)" above
printf("%% Experiment Number : %d \n", MaxExp);
while(ExpNumber <= MaxExp)
{
Init();
int pos[2];
pos[0] = InitPos[0];
pos[1] = InitPos[1];
do{
int direction = (rand() % 4) + 1;
if (moveToDirection(pos, direction))
{
StepCount++;
}
if (pos[1] == 4&&grid[pos[0]][4]==2&&AntsBag==0)
{
CarryFood(pos);
}
if (pos[1] == 0&&grid[pos[0]][0]==0&&AntsBag==2)
{
LeftFood(pos);
}
checkSuccess();
}
while(!Success);
ShowResult();
ExpNumber++;
}
return 0;
}
void Init()
{
Success = false;
StepCount = 0;
AntsBag = 0;
int gridInit[5][5] = {
{0, 0, 0, 0, 2},
{0, 0, 0, 0, 2},
{0, 0, 0, 0, 2},
{0, 0, 0, 0, 2},
{0, 0, 0, 0, 2}
};
for (int i = 0; i < 5; ++i)
{
for (int j = 0; j < 5; ++j)
{
grid[i][j] = gridInit[i][j];
}
}
}
void ShowResult()
{
/*
for (int i = 0; i < 5; ++i)
{
printf("\n");
for (int j = 0; j < 5; ++j)
{
printf("%d ", grid[i][j]);
}
}
*/
printf("%d %d\n", StepCount, ExpNumber);
}
void CarryFood(int * pos)
{
AntsBag = 2;
grid[pos[0]][4] = 0;
}
void LeftFood(int * pos)
{
AntsBag = 0;
grid[pos[0]][0] = 2;
}
bool checkMovability(int * pos, int direction)
{
switch(direction)
{
case 1:
{
if(pos[1]==0){
return false;
}
break;
}
case 2:
{
if (pos[0]==0)
{
return false;
}
break;
}
case 3:
{
if (pos[0]==4)
{
return false;
}
break;
}
case 4:
{
if (pos[1]==4)
{
return false;
}
break;
}
default:
{
printf("Wrong direction input is given!!\n");
return false;
break;
}
}
return true;
}
bool moveToDirection(int * pos, int direction)
{
if ( !checkMovability(pos, direction) )
{
return false;
}
switch(direction){
case 1:
{
pos[1] -= 1;
break;
}
case 2:
{
pos[0] -= 1;
break;
}
case 3:
{
pos[0] += 1;
break;
}
case 4:
{
pos[1] += 1;
break;
}
default:
{
printf("I'm stunned!\n");
return false;
break;
}
}
return true;
}
bool checkSuccess()
{
for (int i = 0; i < 5; ++i)
{
if (grid[i][0] != 2)
{
return false;
}
}
//printf("Success!\n");
Success = true;
return true;
}
并将输出重定向到* .txt文件,并使用以下八度代码找到步数的预期值;
clear
load data.txt
n = data(:,1);
output_precision(15);
mean(n)
%% The actual data
%% milliData1 -> 430.038224000000
%% milliData2 -> 430.031745000000
%% timeTData1 -> 430.029882400000
%% timeTData2 -> 430.019626400000
%% timeUnsigData1 -> 430.028159000000
%% timeUnsigData2 -> 430.009509000000
然而,即使我运行完全相同的代码两次,我也会得到不同的结果,正如您从上面的结果中看到的那样。(注意,我已尝试使用不同的srand(..)输入,原因是我&#39;我要解释一下。)
我认为这是因为我如何为蚂蚁的随机方向生成1-4之间的随机整数,因为据我所知,这个实验的概率分布应该与只要我重复实验大量的时间(在这个特殊情况下5000000次)。
所以我的第一个问题是,我是如何生成随机整数的方法的问题?如果是这样,我们怎样才能克服这个问题,我的意思是我们怎么能够足够随机地生成整数,以便当我们重复相同的实验很多次时,它们之间的期望值小于我得到的这些结果?
答案 0 :(得分:1)
您可以使用类似
的内容std::mt19937 gen{std::random_device{}()};
std::uniform_int_distribution<int> dist{1, 4};
int randNumber = dist(gen);
这会产生更均匀的值分布。