Permutation Descent Counts(递推)

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Permutation Descent Counts(递推)

1968: Permutation Descent Counts Submit Page Summary Time Limit: 1 Sec Memory Limit: 128 Mb Submitted: 123 Solved: 96 Description Given a positive integer, N, a permutation of order N is a one-to-one (and thus onto) function from the…

permutation 2（递推 + 思维）

permutation 2 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Problem Description You are given three positive integers N,x,y.Please calculate how many permutations of 1∼N satisfies the following conditions (We d…

CSU 1968 Permutation Descent Counts

http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1968 题意:对于任一种N的排列A,定义它的E值为序列中满足A[i]>A[i+1]的数的个数.给定N和K(K<=N<=1000),问N的排列中E值为K的个数. 思路: 这道题目和杭电的3664非常像. d[i][j]表示分析到i这个数时的E值为j的个数. 那么如何计算出d[i][j]呢?得根据d[i-1][j]和d[i-1][j-1]递推出来. ①首先考虑d[i-1][j](此时不改变…

【HDOJ6630】permutation 2（递推）

题意:给定x,y,n,有标号从1到n的n个数组,求合法的排列个数模998244353使得 1:p[1]=x 2:p[n]=y 3:相邻两项的差的绝对值<=2 n<=1e5 思路: #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef unsigned int uint; typedef unsigned long long ull; typedef pair<int,int>…

Codeforces Gym10081 A.Arcade Game-康托展开、全排列、组合数变成递推的思想

最近做到好多概率,组合数,全排列的题目,本咸鱼不会啊,我概率论都挂科了... 这个题学到了一个康托展开,有点用,瞎写一下... 康托展开: 适用对象:没有重复元素的全排列. 把一个整数X展开成如下形式: X=a[n]*(n-1)!+a[n-1]*(n-2)!+...+a[i]*(i-1)!+...+a[2]*1!+a[1]*0![1] 其中a[i]为当前未出现的元素中是排在第几个(从0开始),并且0<=a[i]<i(1<=i<=n) 用来求全排列中这个串排第几,康托展开的逆运算就是…

四角递推(CF Working out,动态规划递推)

题目:假如有A,B两个人,在一个m*n的矩阵,然后A在(1,1),B在(m,1),A要走到(m,n),B要走到(1,n),两人走的过程中可以捡起格子上的数字,而且两人速度不一样,可以同时到一个点(哪怕这个点离A很近,离B很远),现在A,B起码相遇于一个点点,相遇点的数字A,B都得不到,求最后A,B总数字之和的最大值 B. Working out Summer is coming! It's time for Iahub and Iahubina to work out, as they both…