题目大意 你开始有一个序列x 它所有项都是0 你有一个操作:x[i]=x[i-1]+1 问你至少几次操作可以让x序列变为给定的a序列 分析 老年人完全不会这种脑子题/kk........ 我们定义b[i]=i-a[i] 所以对于一个连续的上升子序列它的b是相等的 每次取一段想的b序列中最大的a[i]累加即可 代码 #include<iostream> #include<cstdio> #include<cstring> #include<string> #i…

Problem Statement Find the number of the possible tuples of sequences (A0,A1,…,AN) that satisfy all of the following conditions, modulo M: For every i (0≤i≤N), Ai is a sequence of length i consisting of integers between 1 and K (inclusive); For every…

题目大意:考虑 N +1 个数组 {A0,A1,…,AN}.其中 Ai 的长度是 i,Ai 内的所有数字都在 1 到 K 之间. Ai−1 是 Ai 的子序列,即 Ai 删一个数字可以得到 Ai−1. Ai 的字典序大于 Ai−1.输入 N,K,M 问序列个数模 M.解题思路:考虑每次从前面那个状态插入一个数,那么插入位置右边的那个数要严格小于插入的数. fi,j,p 表示当前长度为i,取到数字j,有p个位置可插的方案数.不插在这个位置,那么fi,j,p−1 += fi,j,p(p不为0)还有就…

题意 给出\(n\),\(m\),\(mu\),问有多少个序列组\((A_0,A_1,\dots,A_n)\)满足: 序列\(Ai\)的长度恰好为\(i\) 所有元素均在\([1,m]\) \(A_{i−1}\)是\(A_i\)的子序列 \(A_i\)的字典序大于\(A_{i−1}\) 答案模\(mu\)输出. \(n,k \le 300\) 传送门 思路 又是一道神仙\(dp\) 一个很重要的思路:把数从小往大插入 当我们插入\(i\)时,因为数列中的数都是\(\le i\)的,所以\(i\)…

题目传送门 典型的 Atcoder 风格的计数 dp. 题目可以转化为每次在序列中插入一个 \([1,k]\) 的数,共操作 \(n\) 次,满足后一个序列的字典序严格大于前一个序列,问有多少种操作序列. 显然相同的数可以合并,因为在由相同的数 \(x\) 组成的数段中,在任何位置插入 \(x\),得到的序列都是相同的. 再考虑字典序的问题.你只能序列末尾或者一个 \(<x\) 的数前面插入 \(x\),否则得到的序列的字典序就会 \(\geq\) 原序列的字典序. 但这样问题还是比较棘手,我们…

A - Fairness 如果奇数次是b - a 否则是a - b #include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') //#define ivorysi using na…

Coprime Sequence                                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)                                                                                      Total S…

Description Do you know what is called ``Coprime Sequence''? That is a sequence consists of $n$ positive integers, and the GCD (Greatest Common Divisor) of them is equal to 1. ``Coprime Sequence'' is easy to find because of its restriction. But we ca…

Do you know what is called ``Coprime Sequence''? That is a sequence consists of nn positive integers, and the GCD (Greatest Common Divisor) of them is equal to 1. ``Coprime Sequence'' is easy to find because of its restriction. But we can try to maxi…

Do you know what is called ``Coprime Sequence''? That is a sequence consists of nnpositive integers, and the GCD (Greatest Common Divisor) of them is equal to 1. ``Coprime Sequence'' is easy to find because of its restriction. But we can try to maxim…