HDU 1014 Uniform Generator（模拟和公式）

传送门：

http://acm.hdu.edu.cn/showproblem.php?pid=1014

Uniform Generator

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33120    Accepted Submission(s): 13137

Problem Description

Computer simulations often require random numbers. One way to generate pseudo-random numbers is via a function of the form

seed(x+1) = [seed(x) + STEP] % MOD

where '%' is the modulus operator.

Such a function will generate pseudo-random numbers (seed) between 0 and MOD-1. One problem with functions of this form is that they will always generate the same pattern over and over. In order to minimize this effect, selecting the STEP and MOD values carefully can result in a uniform distribution of all values between (and including) 0 and MOD-1.

For example, if STEP = 3 and MOD = 5, the function will generate the series of pseudo-random numbers 0, 3, 1, 4, 2 in a repeating cycle. In this example, all of the numbers between and including 0 and MOD-1 will be generated every MOD iterations of the function. Note that by the nature of the function to generate the same seed(x+1) every time seed(x) occurs means that if a function will generate all the numbers between 0 and MOD-1, it will generate pseudo-random numbers uniformly with every MOD iterations.

If STEP = 15 and MOD = 20, the function generates the series 0, 15, 10, 5 (or any other repeating series if the initial seed is other than 0). This is a poor selection of STEP and MOD because no initial seed will generate all of the numbers from 0 and MOD-1.

Your program will determine if choices of STEP and MOD will generate a uniform distribution of pseudo-random numbers.

 

Input

Each line of input will contain a pair of integers for STEP and MOD in that order (1 <= STEP, MOD <= 100000).

 

Output

For each line of input, your program should print the STEP value right- justified in columns 1 through 10, the MOD value right-justified in columns 11 through 20 and either "Good Choice" or "Bad Choice" left-justified starting in column 25. The "Good Choice" message should be printed when the selection of STEP and MOD will generate all the numbers between and including 0 and MOD-1 when MOD numbers are generated. Otherwise, your program should print the message "Bad Choice". After each output test set, your program should print exactly one blank line.

 

Sample Input

3 5

15 20

63923 99999

 

Sample Output

3 5 Good Choice

15 20 Bad Choice

63923 99999 Good Choice

 

Source

South Central USA 1996

 

题目意思：

就是给定两个数step, mod然后利用给定的公式求随机数，保证从0到mod减1范围内都均匀出现。
每一个数都出现一次。这样就均等分布了。就打印Good否则就Bad 

 

分析：

有两种做法：

1.直接模拟

枚举产生随机数，然后排序比对

全部匹配就是好的，有一个不是就是坏的

code：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m,i;
    int s[];
    while(cin>>n>>m)
    {
        s[]=;
        for(i=;i<m;i++)
            s[i]=(s[i-]+n)%m;
        sort(s,s+m);
        for(i=;i<m;i++)
            if(s[i]!=i)
                break;
        printf("%10d%10d",n,m);
        if(i==m)
            cout<<"    Good Choice"<<endl<<endl;
        else
            cout<<"    Bad Choice"<<endl<<endl;
    }
    return ;
}

2.公式法

这是网上比较神奇的做法（判断一下两个数是不是互质）

就是比较step和mod的最大公约数是不是1

大佬题解：

本题就是求step和mod如果GCD（最大公约数位1）那么就是Good Choice，否则为Bad Choice

为什么这个结论成立呢？

因为当GCD(step, mod) == 1的时候，那么第一次得到序列：x0, x0 + step, x0 + step…… 那么mod之后，必然下一次重复出现比x0大的数必然是x0+1，为什么呢？

因为(x0 + n*step) % mod； 且不需要考虑x0 % mod的值为多少，因为我们想知道第一次比x0大的数是多少，那么就看n*step%mod会是多少了，因为GCD(step, mod) == 1，那么n*step%mod必然是等于1，故此第一次重复出现比x0大的数必然是x0+1，那么第二次出现比x0大的数必然是x0+2，以此类推，就可得到必然会出现所有0到mod-1的数，然后才会重复出现x0.

当GCD(step, mod) != 1的时候，可以推出肯定跨过某些数了，这里不推了。

然后可以扩展这个结论，比如如果使用函数 x(n) = (x(n-1) * a + b)%mod；增加了乘法因子a，和步长b了；

那么如果是Good Choice，就必然需要GCD(a, mod) == 1，而且GCD(b, mod) == 1；

code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int gcd(int a,int b)//最大公约数
{
    if (b==)
        return a;
    return gcd(b, a%b);
}
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        if(gcd(n,m)==)
            printf("%10d%10d    Good Choice\n\n",n,m);
        else
             printf("%10d%10d    Bad Choice\n\n",n,m);
    }
    return ;
}