Uniform Generator
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29336 Accepted Submission(s): 11694
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.
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.
3 5 Good Choice
15 20 Bad Choice
63923 99999 Good Choice
下面解释一下为什么GCD是正解呢！
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；
这里就偷懒不证明这个扩展结论了,而且证明这个结论需要用到线性模(Congruence)和乘法逆元的知识了。
下面给出AC代码：
#include <bits/stdc++.h>
using namespace std;
int gcd(int a,int b)
{
return b==?a:gcd(b,a%b);
}
int main()
{
int step,mod;
while(scanf("%d%d",&step,&mod)!=EOF)
{
if(gcd(step,mod)==)
printf("%10d%10d Good Choice\n\n",step,mod);
else
printf("%10d%10d Bad Choice\n\n",step,mod);
}
return ;
}