HDU5768Lucky7（中国剩余定理+容斥定理）(区间个数统计)

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 and sent it back to the shore. It is said that ?? used to surrounded by 7…

HDU 5768 Lucky7 (中国剩余定理+容斥)

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…

hdu_5768_Lucky7(中国剩余定理+容斥)

题目链接:hdu_5768_Lucky7 题意: 给你一个区间,问你这个区间内是7的倍数,并且满足%a[i]不等于w[i]的数的个数乍一看以为是数位DP,仔细看看条件,发现要用中国剩余定理,然后容斥一下就出答案了,不过这里在中国剩余定理里面的乘法会有数据爆long long ,所有要写一个高精度乘法,这里卡死很多人. #include <bits/stdc++.h> #define F(i,a,b) for(int i=a;i<=b;i++) using namespace std;…

hdu 5768 Lucky7 中国剩余定理+容斥+快速乘

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…

HDU 5768 Lucky7 （中国剩余定理 + 容斥 + 快速乘法）

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…

【hdu4135】【hdu2841】【hdu1695】一类通过容斥定理求区间互质的方法

[HDU4135]Co-prime 题意给出三个整数N,A,B.问在区间[A,B]内,与N互质的数的个数.其中N<=10^9,A,B<=10^15. 分析容斥定理的模板题.可以通过容斥定理求出[1,n]与x互质的数的个数.方法是先将x进行质因子分解,然后对于每个质因子pi,[1,n]内可以被pi整除的数目为n/pi.可以通过容斥定理解决逆命题,既[1,n]与x不互质的数目.n/p1+n/p2+n/p3-n/p1p2-n/p1p3-n/p2p3+n/p1p2p3.既奇数是加,偶数是减.具体的…

hdu_5213_Lucky(莫队算法+容斥定理)

题目连接:hdu_5213_Lucky 题意:给你n个数,一个K,m个询问,每个询问有l1,r1,l2,r2两个区间,让你选取两个数x,y,x,y的位置为xi,yi,满足l1<=xi<=r1,l2<=y2<=r2,使得x+y=K: 题解:首先,这题没有修改操作,即可以离线,离线区间问题就要想到莫队算法,然后看状态怎么搞,因为要求的答案满足区间的可加性,我们令f(l,r)表示 l到r这个区间满足条件的ans,令F(l1,r1,l2,r2)为在这两个区间内选取的数满足条件的ans,则根…

HDU - 4135 Co-prime 容斥定理

题意:给定区间和n,求区间中与n互素的数的个数, . 思路:利用容斥定理求得先求得区间与n互素的数的个数,设表示区间中与n互素的数的个数, 那么区间中与n互素的数的个数等于.详细分析见求指定区间内与n互素的数的个数容斥原理 AC代码 #include <cstdio> #include <cmath> #include <cctype> #include <bitset> #include <algorithm> #include <cs…

HDU 1695 GCD 欧拉函数+容斥定理 || 莫比乌斯反演

GCD Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4141 Accepted Submission(s): 1441 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y)…

HDU 4135 Co-prime 欧拉+容斥定理

Co-prime Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 935 Accepted Submission(s): 339 Problem Description Given a number N, you are asked to count the number of integers between A and B in…

HDU5213(容斥定理+莫队算法）

传送门题意给出n个数和幸运数k,m次询问,每次询问[l1,r1]和[l2,r2]有多少对数满足x+y=k,x∈[l1,r1],y∈[l2,r2] 分析看到m只有3e4,可以考虑\(m\sqrt{n}\)的莫队算法,具体讲解.首先设f(l,r)表示从l到r满足x+y=k的对数,那么由容斥定理得到,\[f(l1,r1,l2,r2)=f(l1,r2)-f(l1,l2-1)-f(r1+1,r2)+f(r1+1,l2-1)\],那么就可以结合莫队算法离线求出询问结果,具体见代码 trick 代码 #…

题解报告：hdu 4135 Co-prime（容斥定理入门）

Problem Description Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than…

HDU 2841 Visible Trees（容斥定理）

Visible Trees Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1951 Accepted Submission(s): 792 Problem Description There are many trees forming a m * n grid, the grid starts from (1,1). Farm…

HDU 1796How many integers can you find(简单容斥定理)

How many integers can you find Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3315 Accepted Submission(s): 937 Problem Description Now you get a number N, and a M-integers set, you shoul…

How Many Sets I（容斥定理）

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3556 How Many Sets I Time Limit: 2 Seconds Memory Limit: 65536 KB Give a set S, |S| = n, then how many ordered set group (S1, S2, ..., Sk) satisfies S1 ∩ S2 ∩ ... ∩ Sk = ∅. (Si is…

BZoj 2301 Problem b（容斥定理+莫比乌斯反演）

2301: [HAOI2011]Problem b Time Limit: 50 Sec Memory Limit: 256 MB Submit: 7732 Solved: 3750 [Submit][Status][Discuss] Description 对于给出的n个询问,每次求有多少个数对(x,y),满足a≤x≤b,c≤y≤d,且gcd(x,y) = k,gcd(x,y)函数为x和y的最大公约数. 100%的数据满足:1≤n≤50000,1≤a≤b≤50000,1≤c≤d≤50000…

BZOJ2839 : 集合计数 (广义容斥定理)

题目一个有 \(N\) 个元素的集合有 \(2^N\) 个不同子集(包含空集), 现在要在这 \(2^N\) 个集合中取出若干集合(至少一个), 使得它们的交集的元素个数为 \(K\) ,求取法的方案数,答案模 \(1000000007\) . \((1 \le N \le 10^6, 0 \le K \le N)\) 题解又是一道裸的广义容斥定理还没这道题难qwq 广义容斥定理 (二项式反演) : \[\displaystyle b_k = \sum_{i=k}^n \binom…

51nod1284容斥定理

1284 2 3 5 7的倍数基准时间限制:1 秒空间限制:131072 KB 分值: 5 难度:1级算法题给出一个数N,求1至N中,有多少个数不是2 3 5 7的倍数. 例如N = 10,只有1不是2 3 5 7的倍数. Input 输入1个数N(1 <= N <= 10^18). Output 输出不是2 3 5 7的倍数的数共有多少. Input示例 10 Output示例 1经典的容斥定理,公式 |AUBUC|=|A|+|B|+|C|-|A^B|-|A^C|-|B^C|+|A…

HDU 1695 GCD（容斥定理）

GCD Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7529 Accepted Submission(s): 2773 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y…

TOJ 4008 The Leaf Eaters(容斥定理)

Description As we all know caterpillars love to eat leaves. Usually, a caterpillar sits on leaf, eats as much of it as it can (or wants), then stretches out to its full length to reach a new leaf with its front end, and finally "hops" to it by c…

2 3 5 7的倍数 (51Nod - 1284)[容斥定理]

20180604 给出一个数N,求1至N中,有多少个数不是2 3 5 7的倍数. 例如N = 10,只有1不是2 3 5 7的倍数. Input 输入1个数N(1 <= N <= 10^18). Output 输出不是2 3 5 7的倍数的数共有多少. Sample Input 10 Sample Output 1 思路: ⒈如果对数学不是很了解,这道题可以按照常规思路来,只是分得不完,毕竟数据太大会爆掉.(20道测试过了7道) #include<cstdio> #include&…

HihoCoder1655 : 第K小最简真分数（[Offer收割]编程练习赛39）（唯一分解+容斥定理+二分）(不错的数学题)

描述给一个整数N,请你求出以N为分母的最简(既约)真分数中第K小的是多少? 输入两个整数N个K. 对于30%的数据,1 <= N <= 1000000 对于100%的数据,1 <= N <= 10000000000 且 1 <= K <= φ(N).其中φ(N)是欧拉函数,也即1~N中与N互质的数的个数. 输出一个整数表示答案的分子样例输入 100 10 样例输出 23 思路: 显然和欧拉函数关系不大...开始思路已经很接近了.分解质因子,然后二分1到Mid中与…

HDU1796 How many integers can you find【容斥定理】

题目链接: http://acm.hdu.edu.cn/showproblem.php? pid=1796 题目大意: 给你一个整数N.和M个整数的集合{A1.A2.-.Am}.集合内元素为非负数(包括零),求小于N的正整数(1~N-1)中,能被M个整数的集合中随意一个元素整除的正整数个数. 比如N = 12.M = {2,3},在1~N-1中,能被2整除的数为{2,4,6.8.10},能被3整除的数为 {3.6,9}.则所求集合为{2,3,4.6,8,9,10},共7个,则答案为7. 思路:…

Luogu-P1450 [HAOI2008]硬币购物-完全背包+容斥定理

Luogu-P1450 [HAOI2008]硬币购物-完全背包+容斥定理 [Problem Description] 略 [Solution] 上述题目等价于:有\(4\)种物品,每种物品有\(d_i\)个,且每种物品的体积为\(c_i\),问有多少种方法装满容量为\(s\)的背包?可以很容易想到跑多重背包即可,但是发现复杂度为\(O(4V\cdot n)\).不可行. 题目要求的东西也等价于求以下等式有多少组满足条件的解: \[ c_1\cdot x_1+c_2\cdot x_2+c_3\cd…

CRT中国剩余定理 & Lucas卢卡斯定理

数论_CRT(中国剩余定理)& Lucas (卢卡斯定理) 前言又是一脸懵逼的一天. 正文按照道理来说,我们应该先做一个介绍. 中国剩余定理中国剩余定理,Chinese Remainder Theorem,又称孙子定理,给出了一元线性同余方程组的有解判定条件,并用构造法给出了通解的具体形式. 现在有方程组:中国剩余定理指出: 扩展中国剩余定理在一般情况下,要求任两个数互质这个条件太苛刻了,CRT派不上用场,我们需要一个更具普遍性的结论,这就是EX-CRT.虽然是称为EX-CRT,但这个定…

Codeforces Round #330 (Div. 2) B. Pasha and Phone 容斥定理

B. Pasha and Phone Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/595/problem/B Description Pasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of ex…