Description Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she wants her partner to be clever too (although he's not)! Parmida has prepared the following test problem for Pashmak. There is a sequence a that co…

Pashmak and Parmida's problem Time Limit:3000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 459D Description Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she…

题目链接: D. Pashmak and Parmida's problem time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she wants her…

D. Pashmak and Parmida's problem time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she wants her partn…

http://codeforces.com/contest/459/problem/D 题意:给你n个数,然后统计多少组(i,j)使得f(1,i,ai)>f(j,n,aj); 思路:先从左往右统计f(1,i,ai)记录在b[i],然后从右往左统计f(j,n,aj)记录在c[i],然后利用树状数组统计个数即可. #include <cstdio> #include <iostream> #include <cstring> #include <cmath>…

题目链接:http://codeforces.com/contest/459/problem/D 题意:给出数组a,定义f(l,r,x)为a[]的下标l到r之间,等于x的元素数.i和j符合f(1,i,a[i])>f(j,n,a[j]),而且要求i<j,求i和j的种类数. 题解:先预处理一下设map be[a[i]]表示f(1,i,a[i])的值,然后在倒着来设af[a[j]]表示f(j,n,a[j])的值. 然后再用线段树更新一下长度为af[a[j]]的值然后再在树上查询小于be[a[j-1]…

题目链接:http://codeforces.com/contest/459/problem/D 题意: 数列A, ai表示 i-th 的值, f(i,j, x) 表示[i,j]之间x的数目, 问:当 1 <= i < j <= n, 满足 f(1, i, ai) < f(j,n, aj)的对数. 题解:分别求出两组数列  Ii = f(1, i, ai)(1<=i <= n) Jj =  f(j,n, aj)(1 <= j <= n), 然后, 从n开始,…

Codeforces Round #261 (Div. 2)   题意:给出数组A,定义f(l,r,x)为A[]的下标l到r之间,等于x的元素数.i和j符合f(1,i,a[i])>f(j,n,a[j]),求i和j的种类数. 题解:使用树状数组统计小于某数的元素数量. 我们可以先把f(1,i,a[i])和f(j,n,a[j])写出来,观察一下,例如样例1: n=7 A 1 2 1 1 2 2 1 R 4 3 3 2 2 1 1 L 1 1 2 3 2 3 4 其中A为给定的数组,Rj为f(j,n,…

题意:给出数组A,定义f(l,r,x)为A[]的下标l到r之间,等于x的元素数.i和j符合f(1,i,a[i])>f(j,n,a[j]),求有多少对这样的(i,j). 解法:分别从左到右,由右到左预处理到某个下标为止有多少个数等于该下标,用map维护. 然后树状数组更新每个f(j,n,a[j]),预处理完毕,接下来,从左往右扫过去,每次从树状数组中删去a[i],因为i != j,i不能用作后面的统计,然后统计getsum(inc[a[i]]-1), (inc表示从左到右),即查询比此时的a[i]…

题目链接 题意:给出数组A,定义f(l,r,x)为A[]的下标l到r之间,等于x的元素数.i和j符合f(1,i,a[i])>f(j,n,a[j]),求i和j的种类数. 我们可以用map预处理出 f(1, i, a[i]) 和 f(j, n, a[j]) ,记为s1[n], s2[n]. 这样就变成求满足 s1[i] > s[j], i < j 情况的数量了,你会发现跟求逆序对一样 分析: 做题的时候想到过逆序数,但是很快放弃了,还是理解不深刻吧,,,233. 看了这个博客以后才明白这些过…