Disharmony Trees HDU - 3015 One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them. She finds that trees maybe disharmony and the Disharmony Value between two trees is associa…

#include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std; struct node { ll x,h; } Tr[]; ],tr2[]; bool cmp(node s1,node s2) { return s1.x<s2.x; } bool cmp1(node s1,node s2) { return s1.h<s2.h;…

Disharmony Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 663    Accepted Submission(s): 307 Problem Description One day Sophia finds a very big square. There are n trees in the square. T…

Disharmony Trees Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them.…

/* 写完这篇博客有很多感慨,过去一段时间都是看完题解刷题,刷题,看会题解,没有了大一那个时候什么都不会的时候刷题的感觉,这个题做了一天半,从开始到结束都是从头开始自己构思的很有感觉,找回到当初的感觉 */ Disharmony Trees Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 90 Accepted Submission(s):…

C - Visible Trees HDU - 2841 思路 :被挡住的那些点(x , y)肯定是 x 与 y不互质.能够由其他坐标的倍数表示,所以就转化成了求那些点 x,y互质 也就是在 1 - m    1 - n 中找互质的对数,容斥 求一下即可 #include<bits/stdc++.h> using namespace std; #define ll long long #define maxn 123456 bool vis[maxn+10]; ll t,n,m,prime[m…

题意:给你n棵树,每棵树上有两个权值X H 对于X离散化 :3 7 1 5 3 6 -> 2 6 1 4 2 5,对于H一样 然后F = abs(X1-X2)   S=min(H1,H2) 求出每一对F*S的总和 可以看到一边是求每个数与其他数的最小值,一边是求每个数与其他数的差距.因此我们可以排序一边,处理另一边. 我们排序H,因为这样对于固定一个Xi Hi,从小到大每次都是Hi去乘以Xi与剩下的所有X的差的总和. 这样我们就可以使用树状数组维护两个值:每个位置值的个数,每个位置值的总大小,接…

题解:在路边有一行树,给出它们的坐标和高度,先按X坐标排序.记录排名,记为rankx,再按它们的高度排序,记录排名,记为rankh.两颗树i,j的差异度为 fabs(rankx[i]-rankx[j])*min(rankh[i],rankh[j]) 最后求出任异两颗树差异度的和. 题解:首先,需要解决的是min(rh)的问题,对于这个问题,只要离散化之后按rh从大到小排序顺序求解即可,然后用树状数组维护之前出现rx比当前值小的个数与总和,那么同时就可以计算rx比当前大的个数和总和了,那么就可以依…

题意:给出n棵树,给出横坐标x,还有它们的高度h,先按照横坐标排序,则它们的横坐标记为xx, 再按照它们的高度排序,记为hh 两颗树的差异度为 abs(xx[i] - xx[j]) * min(hh[i],hh[j]),求所有的差异度的和 和上面一道题一样,只不过这题是要Min的值,就将h从大到小排序,保证每一个h都是当前最小的 然后维护比当前x小的坐标的个数,当前区间的总和,前面比x小的坐标的和 #include<iostream> #include<cstdio> #inclu…

Problem DescriptionMost of us know that in the game called DotA(Defense of the Ancient), Pudge is a strong hero in the first period of the game. When the game goes to end however, Pudge is not a strong hero any more.So Pudge’s teammates give him a ne…