题意:……应该不用我说了,看起来就很容斥原理,很中国剩余定理…… 方法:因为题目中的n最大是15,使用状态压缩可以将所有的组合都举出来,然后再拆开成数组,进行中国剩余定理的运算,中国剩余定理能够求出同时满足余膜条件的最小整数x,x在(1,M)之间由唯一值,M是各个除数的乘积,所有符合条件的解为ans = x+k*M,可以知道在[1,R]这个区间内,有(M+R-x)/ M个k符合条件,然后在运算中为了防止溢出,所以使用了带膜乘法,就是将乘数转化为二进制,通过位移运算符,在中间过程中不断的取膜(看代…

分析: 因为满足任意一组pi和ai,即可使一个“幸运数”被“污染”,我们可以想到通过容斥来处理这个问题.当我们选定了一系列pi和ai后,题意转化为求[x,y]中被7整除余0,且被这一系列pi除余ai的数的个数,可以看成若干个同余方程联立成的一次同余方程组.然后我们就可以很自然而然的想到了中国剩余定理.需要注意的是,在处理中国剩余定理的过程中,可能会发生超出LongLong的情况,需要写个类似于快速幂的快速乘法来处理. 吐槽:赛场上不会快速乘,导致疯狂WA,唉,还是太年轻 代码: #include…

Lucky7 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5768 Description When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortunately fall into the sea. While it was dying, seven dolphins arched its body an…

http://acm.hdu.edu.cn/showproblem.php?pid=4057 Problem Description Dr. X is a biologist, who likes rabbits very much and can do everything for them. 2012 is coming, and Dr. X wants to take some rabbits to Noah's Ark, or there are no rabbits any more.…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5768 题目大意: T组数据,求L~R中满足:1.是7的倍数,2.对n个素数有 %pi!=ai  的数的个数. 题目思路: [中国剩余定理][容斥原理][快速乘法][数论] 因为都是素数所以两两互素,满足中国剩余定理的条件. 把7加到素数中,a=0,这样就变成解n+1个同余方程的通解(最小解).之后算L~R中有多少解. 但是由于中国剩余定理的条件是同时成立的,而题目是或的关系,所以要用容斥原理叠加删…

http://acm.hdu.edu.cn/showproblem.php?pid=5768 Lucky7 Problem Description   When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortunately fall into the sea. While it was dying, seven dolphins arched its bod…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5768 给你n个同余方程组,然后给你l,r,问你l,r中有多少数%7=0且%ai != bi. 比较明显的中国剩余定理+容斥,容斥的时候每次要加上个(%7=0)这一组. 中间会爆longlong,所以在其中加上个快速乘法(类似快速幂).因为普通的a*b是直接a个b相加,很可能会爆.但是你可以将b拆分为二进制来加a,这样又快又可以防爆. //#pragma comment(linker, "/STACK…

Lucky7 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortunately fall into the sea. While it was d…

Lucky7 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5768 Description When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortunately fall into the sea. While it was dying, seven dolphins arched its body an…

题意:求满足7的倍数,不满足其他条件num % p == a 的num的个数. 思路:利用中国剩余定理我i们可以求出7的倍数,但是多算了不满足约定条件又得减去一个,但是又发现多减了,又得加回来.如此,那么应该应用容斥原理来解决问题.那么就应该是将所有的状态都遍历一下,然后根据1的个数来判断是不是+-号. #include<bits/stdc++.h> using namespace std; typedef long long ll; ; ll n, l, r; ll m[maxn], a[m…