问题描述 给定你一个 \(1 \sim n\) 的排列 \(\{p_i\}\),可进行若干次操作,每次选择两个整数 \(x,y\),交换 \(p_x,p_y\). 请你告诉穰子,用最少的操作次数将给定排列变成单调上升的序列 \(1,2,\ldots,n\),有多少种方式呢?请输出方式数对 \(10^9+9\) 取模的结果. 输入格式 第一行一个整数 \(T\) 代表数据组数. 每一组测试数据,第一行是一个整数 \(n\) 代表排列中的元素个数,第二行 \(n\) 个整数,是这个排列. 输入数据中…

题意 3602 Counting Swaps 0x30「数学知识」例题 背景 https://ipsc.ksp.sk/2016/real/problems/c.html Just like yesterday (in problem U of the practice session), Bob is busy, so Alice keeps on playing some single-player games and puzzles. In her newest puzzle she has…

Counting swaps 给你一个1-n的排列,问用最少的交换次数使之变为递增排列的方案数\(mod\ 10^9+7\),1 ≤ n ≤ 10^5. 解 显然最少的交换次数不定,还得需要找到最小交换次数,而考虑到交换为复杂的过程,考虑状态的性质,所以不难想到画出,+为箭头指向方向 _ _ | + | + 2 1 4 3 + | + | |_| |_| 于是你会发现实际上我们的变换为递增序列,即把所有的环都变成自环,而交换两个数字即拆环,所以不难知道,一个环拆掉的最少的次数为环的大小-1(因为…

正解:数论 解题报告: 传送门! 首先考虑最终的状态是固定的,所以可以知道初始状态的每个数要去哪个地方,就可以考虑给每个数$a$连一条边,指向一个数$b$,表示$a$最后要移至$b$所在的位置 显然每个数只会有一条出边,也只会有一条入边,所以会构成若干条环然后现在的目标就相当于是要通过最少的次数使所有边都变成自环 然后考虑这个交换操作,就相当于是交换两条边的终点 欧克把题目转化完了下面考虑解题 先证明这样一个结论:一个长度为$n$的环要变成$n$个自环至少需要$n-1$步 证明如下: 考虑用数学…

计数套路题?但是我连套路都不会,,, 拿到这道题我一脸蒙彼,,,感谢@poorpool 大佬的博客的指点 先将第\(i\)位上的数字\(p_i\)向\(i\)连无向边,然后构成了一个有若干环组成的无向图,可以知道某个点包含它的有且仅有一个环,因为所有点度数都为2(自环的点度数也是2吧qwq) 那么我们的最终目标就是把这个图转换成有\(n\)个自环的图,相当于把排列排好顺序 考虑对于一个\(m\)个点的环,把它变成\(m\)个自环至少需要\(m-1\)步(相当于排列\((2,3...m,1)\)变…

题解 #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> #define ll long long using namespace std; const int MOD = 1e9 + 9, MAXN = 100005; int T, n, num[MAXN], head[MAXN], nume, id[MAX…

题目链接 题解 首先,对于每个\(i\)向\(a[i]\)连边. 这样会连出许多独立的环. 可以证明,交换操作不会跨越环. 每个环内的点到最终状态最少交换步数是 \(环的大小-1\) 那么设\(f[i]\)表示环大小为\(i\)的方案数 则 \[ f[i] = \sum_{x+y=i}f[x] * f[y] * g(x,y) * (^{i-1}_{x-1}) \] 其中 \[ g(x,y)=\{^{\frac{x+y}{2}{(x+y为偶数且x=y)}}_{x+y(else)} \] 打标可以发…

第一道 A 掉的严格意义上的组合计数题,特来纪念一发. 第一次真正接触到这种类型的题,给人感觉好像思维得很发散才行-- 对于一个排列 \(p_1,p_2,\dots,p_n\),对于每个 \(i\) 向 \(p_i\) 连一条边,可以发现整个构成了一个由若干环组成的图,目标是将这些环变为自环. 引理:把长度为 \(n\) 的环变为 \(n\) 个自环,最少交换次数为 \(n-1\). 用归纳法证,对于当前情况,任意一次交换都将其拆为两个环,由淘汰赛法则可知引理成立. 记 \(F_n\) 表示在最…

组合计算的性质: C(n,m)= m! / (n!(m-n)!) C(n,m)=C(m-n,m); C(n,m)=C(n,m-1)+C(n-1,m-1); 二项式定理:(a+b)^n=sigema(k=0~n) C(k,n)*a^k*b^(n-k) lucas定理:C(n,m)≡C(n%p,m%p)*C(n/p,m/p)  (mod p) catalan数: Cat(n)=C(n,2n)/n+1  Cat(n)=Cat(n-1)*(4n-2)/(n+1) 计算系数 通过二项式定理变形其实就是求C…

在上篇,我了解了基数的基本概念,现在进入Linear Counting算法的学习. 理解颇浅,还请大神指点! http://blog.codinglabs.org/articles/algorithms-for-cardinality-estimation-part-ii.html 它的基本处理方法和上篇中用bitmap统计的方法类似,但是最后要用到一个公式: 说明:m为bitmap总位数,u为0的个数,最后的结果为n的一个估计,且为最大似然估计(MLE). 那么问题来了,最大似然估计是什么东东…

来之不易的2017第一发ac http://poj.org/problem?id=2386 Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31474   Accepted: 15724 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is repr…

ZOJ3944 People Counting ZOJ3939 The Lucky Week 1.PeopleConting 题意:照片上有很多个人,用矩阵里的字符表示.一个人如下: .O. /|\ (.) 占3*3格子,句号“.”为背景.没有两个人完全重合.有的人被挡住了一部分.问照片上有几个人. 题解: 先弄个常量把3*3人形存起来,然后6个部位依次找,比如现在找头,找到一个头,就把这个人删掉(找这个人的各个部位,如果在该部位位置的不是这个人的身体,就不删),删成句号,疯狂找就行了. 代码:…

#include <stdio.h> #include <malloc.h> #define MAX_STACK 10 ; // define the node of stack typedef struct node { int data; node *next; }*Snode; // define the stack typedef struct Stack{ Snode top; Snode bottom; }*NewStack; void creatStack(NewSt…

1004. Counting Leaves (30)   A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100,…

Problem Figure 2. The Hamming distance between these two strings is 7. Mismatched symbols are colored red. Given two strings ss and tt of equal length, the Hamming distance between ss and tt, denoted dH(s,t)dH(s,t), is the number of corresponding sym…

Problem A string is simply an ordered collection of symbols selected from some alphabet and formed into a word; the length of a string is the number of symbols that it contains. An example of a length 21 DNA string (whose alphabet contains the symbol…

// uva 11401 Triangle Counting // // 题目大意: // // 求n范围内,任意选三个不同的数,能组成三角形的个数 // // 解题方法: // // 我们设三角巷的最长的长度是c(x),另外两边为y,z // 则由z + y > x得, x - y < z < x 当y = 1时,无解 // 当y = 2时,一个解,这样到y = x - 1 时 有 x - 2个 // 解,所以一共是0,1,2,3....x - 2,一共(x - 2) * (x - 1…

当我们在使用JSONKit处理数据时,直接将文件拉进项目往往会报这两个错“JSONKit   does not support Objective-C Automatic Reference Counting(ARC)”,“ARC forbids Objective-C objects in struct”,这是由于JSONKit库未更新,不支持ARC机制.我们可以参照如下步骤解决:…

有使用JSonKit的朋友,如果遇到“JSonKit does not support Objective-C Automatic Reference Counting(ARC)”这种情况,可参照如下方法: 点击项目根目录->targets->Build Phases->JSONKit.m->添加“-fno-objc-arc”字段,在运行就OK了.…

Lake Counting Time Limit: 1000MS     Memory Limit: 65536K Total Submissions: 17917     Accepted: 9069 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <=…

[codeforces 339]E. Three Swaps 试题描述 Xenia the horse breeder has n (n > 1) horses that stand in a row. Each horse has its own unique number. Initially, the i-th left horse has number i. That is, the sequence of numbers of horses in a row looks as foll…

Boring Counting Time Limit: 3000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述     In this problem you are given a number sequence P consisting of N integer and Pi is the ith element in the sequence. Now you task is to answer a list of queries, for each…

Boring counting Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others) Total Submission(s): 2811    Accepted Submission(s): 827 Problem Description In this problem we consider a rooted tree with N vertices. The vertices a…

Counting Squares Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1885    Accepted Submission(s): 946 Problem Description Your input is a series of rectangles, one per line. Each rectangle is sp…

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2610 Boring Counting Time Limit: 3000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述     In this problem you are given a number sequence P consisting of N integer and Pi is the ith ele…

转载地址:http://blog.csdn.net/xbl1986/article/details/7216668 Xcode是Version 4.2 Build 4D151a 根据Objective-c 2.0程序设计上的旧版本的代码会发生NSAutoreleasePool' is unavailable: not available in automatic reference counting mode的错误 需要手动关闭工程中ARC 工程中 Build Settings--->Apple…

Boring counting Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others) Problem Description In this problem we consider a rooted tree with N vertices. The vertices are numbered from 1 to N, and vertex 1 represents the root…

Counting Offspring Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description You are given a tree, it’s root is p, and the node is numbered from 1 to n. Now define f(i) as the number of nodes whose numbe…

题目描述 "But baby, I've been, I've been praying hard,     Said, no more counting dollars     We'll be counting stars" Grandpa Shaw loves counting stars. One evening he was sitting in his armchar, trying to figure out this problem. Consider the sky…

Counting Bits Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example:For num = 5 you should return [0,1,1,2,1,2]. Follow up…