hdu3664 Permutation Counting 题目传送门 题意: 在一个序列中,如果有k个数满足a[i]>i:那么这个序列的E值为k,问你 在n的全排列中,有多少个排列是恰好是E值为k的序列? 思路: 定义dp[i][j]: 在 i 的全排列中,E值为j的个数:则从i转移到i+1时,有三种情况: 1)把i+1加到最后,E值不变: 2)把i+1与那些已经满足a[i]>i的数交换,E值不变: 3)把i+1与那些不满足a[i]>i的数交换,E值加一. 根据上面得到的转移方程为: d…

Permutation Counting Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1487    Accepted Submission(s): 754 Problem Description Given a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-val…

Permutation Counting Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1171    Accepted Submission(s): 587 Problem Description Given a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-va…

Permutation Counting Given a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-value as the amount of elements where ai > i. For example, the E-value of permutation {1, 3, 2, 4} is 1, while the E-value of {4, 3, 2, 1} is 2. You are requested…

Discription Given a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-value as the amount of elements where ai > i. For example, the E-value of permutation {1, 3, 2, 4} is 1, while the E-value of {4, 3, 2, 1} is 2. You are requested to find h…

题意:给一个 n,求在 n 的所有排列中,恰好有 k 个数a[i] > i 的个数. 析:很明显是DP,搞了好久才搞出来,觉得自己DP,实在是太low了,思路是这样的. dp[i][j]表示 i 个排列,恰好有 j 个数,dp[i][j] = dp[i-1][j] * (j+1) + dp[i-1][j-1] * (i-j).这是状态转移方程. 为什么是这样呢,dp[i-1][j] * (j+1) 意思是,你前i-1个已经凑够 j 个了,那么我把 i 可以去替换这个 j 个任何一个,再加上,把这…

题意: 给你一个n和一个长度为n-1的由0/1构成的b序列 你需要从[1,n]中构造出来一个满足b序列的序列 我们设使用[1,n]构成的序列为a,那么如果ai>ai+1,那么bi=1,否则bi=0 问你你可以构造出来多少满足b序列的序列a 代码: 看官方题解 代码: #include<stack> #include<queue> #include<map> #include<cstdio> #include<cstring> #includ…

Problem J Permutation Counting Dexter considers a permutation of first N natural numbers good if it doesn't have x and x+1 appearing consecutively, where (1 ≤ x < N)  For example, for N=3 , all goodpermutations are: 1. {1, 3, 2} 2.{2, 1, 3} 3.{3, 2,…

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). The replaceme…

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…