题目链接:http://lightoj.com/volume_showproblem.php?problem=1319 In 'MonkeyLand', there is a traditional game called "Bamboo Climbing". The rules of the game are as follows: ) There are N monkeys who play this game and there are N bamboos of equal he…

题目链接:https://vjudge.net/contest/28079#problem/U 题目大意:给你n(n<12)行,每行有pi,ri,求一个数ans满足ans%pi=ri(i从1~n). 解题思路:就是解同余方程组:比如第一组样例: 3 5 4 7 6 11 3 就是要我们求一个x满足下列同余方程组: x≡4(mod5) x≡6(mod7) x≡11(mod3) 然后自己看这里吧写了解释http://www.cnblogs.com/fu3638/p/7453419.html,是CRT…

Monkey Tradition Time Limit: 2000MS   Memory Limit: 32768KB   64bit IO Format: %lld & %llu Submit Status Description In 'MonkeyLand', there is a traditional game called "Bamboo Climbing". The rules of the game are as follows: 1)       There…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1319 题意:有 n 个猴子,n 棵树,树的高度为 L ,每个猴子刚开始的时候都在树的底部,后来往上跳,每次跳的距离是pi,最后不能跳到树上面所以最后会有个到顶端的距离ri,求L的最小值: 典型的模板题: 中国剩余定理详解:http://www.cnblogs.com/zhengguiping--9876/p/5319813.html #include<stdio.h> #inclu…

二进制枚举+容斥原理+中国剩余定理 #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<cmath> using namespace std; #define MAXN 20 typedef long long LL; int n; int s[MAXN]; LL a[MAXN], m[MAXN]; //a是余数,m是除数 LL ex…

我理解的中国剩余定理的含义是:给定一个数除以一系列互素的数${p_1}, \cdots ,{p_n}$的余数,那么这个数除以这组素数之积($N = {p_1} \times  \cdots  \times {p_n}$)的余数也确定了,反之亦然. 用表达式表示如下: \[\begin{array}{l}x \equiv {a_1}(\bmod {p_1})\\{\rm{     }} \vdots \\x \equiv {a_n}(\bmod {p_n})\end{array}\] 那么任何满足…

题目链接: http://www.51nod.com/onlineJudge/user.html#!userId=21687 题意: 中文题诶~ 思路: 本题就是个中国剩余定理模板题,不过模拟也可以过,而且时间复杂度嘛~ 我们可以知道gcd得出两个数的最大公约在最坏的情况下(a, b是相邻的两个斐波拉契数)是O(logn)的, 同理可以知道exgcd也是O(lgn)时间复杂度,因此中国剩余定理时间复杂度是O(nlogn); 而模拟的话最坏的情况下需要O(n*x)的时间~本题两种算法都是15ms.…

Unknown Treasure Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2389    Accepted Submission(s): 885 Problem Description On the way to the next secret treasure hiding place, the mathematician…

/* 中国剩余定理可以描述为: 若某数x分别被d1..….dn除得的余数为r1.r2.….rn,则可表示为下式: x=R1r1+R2r2+…+Rnrn+RD 其中R1是d2.d3.….dn的公倍数,而且被d1除,余数为1:(称为R1相对于d1的数论倒数) R1 . R2 . … . Rn是d1.d2.….dn-1的公倍数,而且被dn除,余数为1: D是d1.d2.….的最小公倍数: R是任意整数(代表倍数),可根据实际需要决定: 且d1..….必须互质,以保证每个Ri(i=1,2,…,n)都能求…

Biorhythms Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 110991   Accepted: 34541 Description Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical,…