链接:pid=4790">http://acm.hdu.edu.cn/showproblem.php?pid=4790 意:从[a.b]中随机找出一个数字x,从[c.d]中随机找出一个数字y.给出p.m,假设(x+y)%p==m则算成功,问成功的概率是多少. 思路:[a.b]中连续p个数.[c,d]中连续p个数.用这2*p个数进行组合能找到p种的成功组合(详细不证),所以找到[a.b]中p循环的个数x1,[c,d]中p循环的个数y1,则它们组成的成功组合数为p*x1*y1. 然后是处理边界…

Just Random Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 87    Accepted Submission(s): 34 Problem Description Coach Pang and Uncle Yang both love numbers. Every morning they play a game with…

题意:在[a,b]  [c,d] 之间,和模p等于m的对数 详见代码 #include <stdio.h> #include <algorithm> #include <string.h> #include<cmath> #define LL long long using namespace std; int T; LL a,b,c,d,p,m; LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a; } LL…

Problem Description Coach Pang and Uncle Yang both love numbers. Every morning they play a game with number together. In each game the following will be done: . Coach Pang randomly choose a integer x in [a, b] with equal probability. . Uncle Yang ran…

/** 大意: 给定[a,b],[c,d] 在这两个区间内分别取一个x,y 使得 (x+y)%p = m 思路:res = f(b,d) -f(b,c-1)-f(a-1,d)+f(a-1,c-1); f(b,d ) 表示在[0,b],[0,d] 之间有多少个符合上述要求的数 1.将[0,b] 分为两部分, b/p 和 b%p 能整除p的[0,(b/p)*p] 和[(b/p)*p+1,b ] 同理[0,d]也可以这样分, 这样对于[0,b] [0,d ] 分别有两种情况,则一共有四种情况. a.…

思路:对于a<=x<=b,c<=y<=d,满足条件的结果为ans=f(b,d)-f(b,c-1)-f(a-1,d)+f(a-1,c-1). 而函数f(a,b)是计算0<=x<=a,0<=y<=b满足条件的结果.这样计算就很方便了. 例如:求f(16,7),p=6,m=2. 对于x有:0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 对于y有:0 1 2 3 4 5 0 1 很容易知道对于xy中的(0 1 2 3 4 5)对满足条件的数目为p…

题目:hdoj 4790 Just Random 题意:给你两个闭区间[a,b],[c,d],分别从中等可能的跳出 x 和 y ,求(x+y)%p == m的概率 分析: 假如是[3,5] [4,7]   p = 2 . m = 1: 则全部的和 7 8 9 10 8 9 10 11 9 10 11 12 1 2 3 3 2 1 后面一行出现次数,能够发现能够分成三部分.第一部分递增的等差数列,第二部分值都相等,第三部分等差数列 然后用等差数列求和公式求和就ok AC代码: #include<i…

Just Random Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3932 Accepted Submission(s): 1276 Problem Description Coach Pang and Uncle Yang both love numbers. Every morning they play a game with n…

Coach Pang and Uncle Yang both love numbers. Every morning they play a game with number together. In each game the following will be done:  1. Coach Pang randomly choose a integer x in [a, b] with equal probability.  2. Uncle Yang randomly choose a i…

Maximum Random Walk Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 756    Accepted Submission(s): 419 三维dp,一维的话根本没有办法开展,二维的话没办法保存当前位置或者最远位置,所以只能用三维的. 看不懂滚动数组之类的操作,只能傻傻的写. 具体内容在代码里标注了,三重循环,从i,j,…