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…

Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 75    Accepted Submission(s): 38 Problem Description Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him to do…

G(x) Time Limit: 2000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 184    Accepted Submission(s): 44 Problem Description For a binary number x with n digits (AnAn-1An-2 ... A2A1), we encode it as Where ""…

题意:问生成树里能不能有符合菲波那切数的白边数量 思路:白边 黑边各优先排序求最小生成树,并统计白边在两种情况下数目,最后判断这个区间就可以.注意最初不连通就不行. #include <stdio.h> #include <algorithm> #include <string.h> #include<cmath> #define LL long long using namespace std; int t,n,m; int tot; ]; struct…

这是一道5Y的题目 有坑的地方我已在代码中注释好了 QAQ Ps:模拟题还是练的太少了,速度不够快诶 //#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h> #include <iostream> #include <climits> #include <cstring> #include <cmath> #inclu…

题意很简单,就是两个大矩阵相乘,然后求乘积. 用 Strassen算法 的话,当N的规模达到100左右就会StackOverFlow了 况且输入的数据范围可达到800,如果变量还不用全局变量的话连内存开辟都开不出来 #pragma comment(linker, "/STACK:16777216") #include <iostream> #include <stdio.h> #define ll long long using namespace std; ;…

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…

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.…