• hdu 1014 Uniform Generator


    Uniform Generator

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 21975    Accepted Submission(s): 8634


    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
     
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    一道水题,不过题目很长,其实理解了代码很容易,注意输出格式。
     
    题意: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 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 就是枚举产生随机数,然后排序在比对,注意下输出的格式就OK了。
     
    附上代码:
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 using namespace std;
     6 
     7 int random[100005];
     8 
     9 int main()
    10 {
    11     int step,mod,i,flag;
    12     while(~scanf("%d%d",&step,&mod))
    13     {
    14         flag=1;
    15         random[0]=0;
    16         for(i=1; i<mod; i++)
    17         {
    18             random[i]=(random[i-1]+step)%mod;   //第一个随机数为0,后面一次出现mod-1个随机数
    19         }
    20         sort(random,random+mod);    //排序
    21         for(i=0; i<mod; i++)
    22             if(random[i]!=i)     //如果有重复的数字出现,则不满足题意
    23             {
    24                 flag=0;
    25                 break;
    26             }
    27         if(flag)
    28             printf("%10d%10d    Good Choice
    
    ",step,mod);
    29         else
    30             printf("%10d%10d    Bad Choice
    
    ",step,mod);
    31     }
    32     return 0;
    33 }

    其实这道题还有一种解释(看了网上大神的博客) 判断两个数是否互质,互质则输出Good,否则输出Bad。

    附上另一种代码:

     1 #include <iostream>
     2 #include <stdio.h>
     3 using namespace std;
     4 int main()
     5 {
     6     int a, b, c, t, m, n;
     7     while(~scanf("%d%d",&a,&b))
     8     {
     9         m=a;
    10         n=b;
    11         if(a<b)  //保证a一定比b大
    12         {
    13             t=a;
    14             a=b;
    15             b=t;
    16         }
    17         while(b!=0)  //辗转相除求最大公约数
    18         {
    19             c=a%b;
    20             a=b;
    21             b=c;
    22         }
    23         if(a==1)
    24             printf("%10d%10d    Good Choice
    
    ",m,n);
    25         else
    26             printf("%10d%10d    Bad Choice
    
    ",m,n);
    27     }
    28     return 0;
    29 }
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  • 原文地址:https://www.cnblogs.com/pshw/p/4776247.html
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