/*POJ 1064 *题目大意:有N条绳子,他们的长度分别为Li,如果从他们中切割K条长度相同的绳子的话,这K条绳子每条最长能有多长?答案保留到小数点后2位 *算法分析:这个问题用二分搜索可以非常容易的求得答案.让我们套用二分搜索的模型试着解决这个问题. * 令:条件C(x):=可以得到K条长度为x的绳子 * 则问题变成了求满足C(x)条件的最大的x.在区间初始化时, 只需使用充分大的数INF作为上界即可 * 现在的问题是是否可以高效的判断C(x).由于长度为Li的绳子最多可以切出floor(…

题目传送门 /* 题意:n条绳子问切割k条长度相等的最长长度 二分搜索:搜索长度,判断能否有k条长度相等的绳子 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; double w[MAXN]; int n, k; int check(double len…

Given a binary tree, find the largest subtree which is a Binary Search Tree (BST), where largest means subtree with largest number of nodes in it. Note:A subtree must include all of its descendants.Here's an example: 10 / \ 15 / \ \ 1 8 7 The Largest…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7156    Accepted Submission(s): 3318 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7493    Accepted Submission(s): 3484 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,…

题目传送门 题意:问使得sum (k^i) = n || n -1 (1 <= i <= r) 的min (r*k)组合的r和k  分析:r的最大不会超过40,枚举r,二分搜索k.注意会爆long long,所以上界需要优化.r = 2开始上界就小于1e6,cyd将后面的范围也求出来了,其实1e6就够用了. 这水题卡了我好久,没有很好分析题目,做不出来就有种无力感,开始烦躁起来.还是题目做得少了,如果这种题做多了,可能看一眼就能做出来了. /**************************…

还是写一下,二分搜索好了 这道题开数组比较坑... 二分,需要注意边界问题,例如:左闭右闭,左闭右开,否则查找不到or死循环 先上AC代码 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct Ch{ char a[12]; char b[12]; }; int cmp(const void *aa,const…

Description Demy has n jewels. Each of her jewels has some value vi and weight wi. Since her husband John got broke after recent financial crises, Demy has decided to sell some jewels. She has decided that she would keep k best jewels for herself. Sh…

题意:给出三种操作 0 e:将e放入容器中 1 e:将e从容器中删除,若不存在,则输出No Elment! 2 a k:搜索容器中比a大的第k个数,若不存在,则输出Not Find! 思路:树状数组+二分搜索,具体见代码吧. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> /* AC 树状数组+二分搜索 题意:给出三种操作 0 e:将e放入容器中…

Yougth的最大化 时间限制:1000 ms | 内存限制:65535 KB 难度:4 描述 Yougth现在有n个物品的重量和价值分别是Wi和Vi,你能帮他从中选出k个物品使得单位重量的价值最大吗? 输入 有多组测试数据 每组测试数据第一行有两个数n和k,接下来一行有n个数Wi和Vi. (1<=k=n<=10000) (1<=Wi,Vi<=1000000) 输出 输出使得单位价值的最大值.(保留两位小数) 样例输入 3 2 2 2 5 3 2 1 样例输出 0.75 来源 Yo…