HDU1014
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Uniform Generator
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28915 Accepted Submission(s): 11513
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.
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
3 515 2063923 99999
Sample Output
3 5 Good Choice 15 20 Bad Choice 63923 99999 Good Choice//公式题 利用公式seed(x+1) = [seed(x) + STEP] % MOD,计算在给定的step 和 mod 中,计算的结果是否能够取便mod 以内的所有整数,为了防止最后的计算超时 可以将公式改写为 seed(x+1) = [seed(x)%MOD + STEP%MOD] % MOD,或者寻找给定的step 和 mod 是否有公约数,有的话为bad 没有的话为good//利用公式#include <iostream>#include <cstdio>#include <algorithm>#include <iomanip>#include <cmath>#include <time.h>using namespace std;int step,mod;bool caculate(){ int i; result[0] = 0; for(i = 0;i < mod-1; ++i) result[i+1] = (result[i] + step) % mod; long len = 0; sort(result,result + mod); len = unique(result, result + mod) - result; if(len == mod) return 1; else return 0;}int main(int argc, const char * argv[]) { while(scanf("%d %d",&step,&mod) != EOF ) { printf("%10d%10d ",step,mod); printf(caculate()==1?"Good Choice\n\n":"Bad Choice\n\n"); } return 0;}//利用最大公约数--更简单#include <iostream>#include <cstdio>#include <algorithm>#include <iomanip>#include <cmath>#include <time.h>using namespace std;int step,mod;int gcd(int a,int b){ if(a % b == 0) return b; else return gcd(b,a%b);}int main(int argc, const char * argv[]) { while(scanf("%d %d",&step,&mod) != EOF ) { printf("%10d%10d ",step,mod); printf(gcd(step,mod)==1?"Good Choice\n\n":"Bad Choice\n\n"); } return 0;}
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