1014 Uniform Generator】的更多相关文章

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 simulati…
Uniform Generator Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29336    Accepted Submission(s): 11694 Problem Description Computer simulations often require random numbers. One way to generat…
Uniform Generator Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24381    Accepted Submission(s): 9642 Problem Description Computer simulations often require random numbers. One way to generat…
Uniform Generator Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32990    Accepted Submission(s): 13081 Problem Description Computer simulations often require random numbers. One way to generat…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1014 题目意思:给出 STEP 和 MOD,然后根据这个公式:seed(x+1) = [seed(x) + STEP] % MOD,问是否在一个周期里可以产生 0 - mod-1 的数.可以的话输出 "Good Choice", 否则输出 "Bad Choice". 好久以前留下来的问题了,以前觉得题目意思又长,以为是很难的题目......今天看<短码之美>…
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-…
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28661    Accepted Submission(s): 11402 Problem Description Computer simulations often require random numbers. One way to generate pseudo-random nu…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1014 解题思路: 1. 把题目意思读懂后,明白会输入两个数,然后根据题中的公式产生一系列伪随机数,看这些数是不是能够包含0~MOD-1.如果产生不了就输出“Good Choice”,否则输出“Bad Choice”. 2. 全部假使x从0开始,设STEP为a,MOD为b.如果说a,b存在倍数关系,即假设最小存在2倍关系,那么就会有b=2a.x初始为0,第一步之后为a,第二步之后为0,之后a.0交替出…
http://acm.hdu.edu.cn/showproblem.php?pid=1014 题目的英文实在是太多了 ,搞不懂. 最后才知道是用公式seed(x+1) = [seed(x) + STEP] % MOD 来计算随机数 ,问是否满足随机数. 初级版本: 思路:把所有的用该公式计算出来的数(存在数组中)都遍历出来,然后排序.由于数字是在0 到mod-1 之间,所以数组的下标必然等于数组的值,有一个不等于,就是bad chioce #include <stdio.h> #include…
找到规律之后本题就是水题了.只是找规律也不太easy的.证明这个规律成立更加不easy. 本题就是求step和mod假设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? 由于当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,必定下一次反复出现比x0大的数必定是x0+1,为什么呢? 由于(x0 + n*step) % mod. 且不须要考虑x0 % mo…
摘取于http://blog.csdn.net/kenden23/article/details/37519883: 找到规律之后本题就是水题了,不过找规律也不太容易的,证明这个规律成立更加不容易. 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…
#include<iostream> #include <cstdio> #include <cstring> using namespace std; int main(void) { int s,m,i; /*int seed=0; int a[10005]={0}; //memset(a,0,sizeof(a));*/ while(scanf("%d%d",&s,&m)!=EOF); { int seed=0; int a[10…
Uniform Generator Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35190 Accepted Submission(s): 14002 Problem Description Computer simulations often require random numbers. One way to generate pseu…
Uniform Generator Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21176 Accepted Submission(s): 8294 Problem Description Computer simulations often require random numbers. One way to generate pseu…
Uniform Generator  Computer simulations often require random numbers. One way to generate pseudo-random numbers is via a function of the form where ``  " is the modulus operator. Such a function will generate pseudo-random numbers (seed) between 0 an…
http://acm.hdu.edu.cn/showproblem.php?pid=1014 给出式子seed(x+1) = [seed(x) + STEP] % MOD seed初始为0,给出STEP和MOD的值 问seed能否取到0~(MOD - 1)之间的所有值 简单模拟 # include <stdio.h> int main() { int Step, Mod, i, Seed, Flag[100005]; while(scanf("%d %d",&Ste…
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…
Computer simulations often require random numbers. One way to generate pseudo-random numbers is via a function of the form where ``  " is the modulus operator. Such a function will generate pseudo-random numbers (seed) between 0 and MOD-1. One proble…
本文参考资料:http://hi.baidu.com/bnjyjncwbdbjnzr/item/1f997cfdd225d5d143c36a58 题意:一个生成随机数的函数, Seed[x+1] = ( seed[x] + STEP ) % MOD 输入step和mod,问能否生成0~MOD-1之间所有的数,是Good Choice,否则Bad Choice 题意其实就是:给出S和M,求0*S%M,1*S%M,2*S%M......(M-1)*S%M能否组成一个集合包含0.1....M-1:(这…
题意:根据这个式子来递推求得每个随机数x,step和mod给定,seed(0)=0.如果推出来的序列是mod个不重复的数字(0~mod-1)则打印good,否则bad(因为不能产生所有的数). 思路: 用一个数组记录所产生过的数,当出现数字已记录过时,判断是否个数为mod个.若是就返回good. #include <bits/stdc++.h> #define LL long long using namespace std; ; bool ans[N]; int main() { //fre…
题意 给你公式seed(x+1) = [seed(x) + STEP] % MOD ,输入step和mod, 问你是否可以从第一项0,算到mod,它们是否都不同 是 good choice 否则 bad choice 分析 枚举过去 code #include<iostream> #include<string.h> #include<algorithm> #include<string> using namespace std; #define ll lo…
Uniform Generator Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u   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…
Uniform Generator Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26820 Accepted Submission(s): 10629 Problem Description Computer simulations often require random numbers. One way to generate pse…
杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…
acm之pku题目分类 对ACM有兴趣的同学们可以看看 DP:  1011   NTA                 简单题  1013   Great Equipment     简单题  1024   Calendar Game       简单题  1027   Human Gene Functions   简单题  1037   Gridland            简单题  1052   Algernon s Noxious Emissions 简单题  1409   Commun…
各种杂题,水题,模拟,包括简单数论. 1001 A+B 1002 A+B+C 1009 Fat Cat 1010 The Angle 1011 Unix ls 1012 Decoding Task 1019 Grandpa's Other Estate 1034 Simple Arithmetics 1036 Complete the sequence! 1043 Maya Calendar 1054 Game Prediction 1057 Mileage Bank 1067 Rails 10…
以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight Moves1101 Gamblers1204 Additive equations 1221 Risk1230 Legendary Pokemon1249 Pushing Boxes 1364 Machine Schedule1368 BOAT1406 Jungle Roads1411 Annive…
转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…
例10        最大公约数 问题描述 有三个正整数a,b,c(0<a,b,c<10^6),其中c不等于b.若a和c的最大公约数为b,现已知a和b,求满足条件的最小的c. 输入数据 第一行输入一个n,表示有n组测试数据,接下来的n行,每行输入两个正整数a,b. 输出格式 输出对应的c,每组测试数据占一行. 输入样例 2 6 2 12 4 输出样例 4 8 (1)编程思路. 利用转辗相除法求两个整数的最大公约数.例如,求整数m=48,n=18两个数的最大公约数的方法如左图所示. 具体做法是:…
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