题目传送门 /* 二分搜索:式子两边取对数,将x提出来,那么另一边就是一个常数了,函数是:lnx/x.二分搜索x,注意要两次 */ #include <cstdio> #include <algorithm> #include <cmath> using namespace std; const double e = exp (1.0); double cal(double x) { return log (x) / x; } int main(void) { //HD…

题目传送门 /* 二分搜索:枚举高度,计算体积与给出的比较. */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; ; const double PI = acos (-1.0); double r, R, H, V; double x; double ca…

公式转化+二分答案 首先,把题目中给的等式转化一下,变成了这个样子. 等式右边的值是可以求出来的. ln(x)/x的大致图像是这样的 那么只要对[0,e]和[e,+∞]分别进行二分答案即可. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; double Y,k; const double e=2.71828182845904…

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,…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12766    Accepted Submission(s): 5696 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y…

DP: DP[len][k][i][j] 再第len位,第一个数len位为i,第二个数len位为j,和的第len位为k 每一位能够从后面一位转移过来,能够进位也能够不进位 A Famous Equation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 500    Accepted Submission(s): 147 Proble…

Problem Description Now,given the equation 8x^4 + 7x^3 + 2x^2 + 3x + 6 == Y,can you find its solution between 0 and 100; Now please try your lucky. Input The first line of the input contains an integer T(1<=T<=100) which means the number of test cas…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=2199 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13475    Accepted Submission(s): 6010 Problem Description Now,given the equation 8*x^4 + 7*x^3…

题目传送门 /* 二分搜索:在0-1e6的范围找到最小的max (ai - bi),也就是使得p + 1 <= a[i] + c or a[i] - c 比赛时以为是贪心,榨干智商也想不出来:( */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <iostream> using namespace std; ; con…

题意:给出一个数让你求出等于这个数的x 策略:如题. 由于整个式子是单调递增的.所以能够用二分. 要注意到精度. 代码: #include <stdio.h> #include <string.h> #include <math.h> #define eps 1e-10 #define f(x) 8*pow(x, 4) + 7*pow(x, 3) + 2*pow(x, 2) + 3*x int main() { int t; double n; scanf("…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5667    Accepted Submission(s): 2681http://acm.hdu.edu.cn/showproblem.php?pid=2199 Problem Description Now,given the…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…

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…

B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following proble…

 FZU 2102   Solve equation Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  Practice Description You are given two positive integers A and B in Base C. For the equation: A=k*B+d We know there always existing many non-nega…

Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…

卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…

题意: 有1~9数字各有a1, a2, -, a9个, 有无穷多的+和=. 问只用这些数字, 最多能组成多少个不同的等式x+y=z, 其中x,y,z∈[1,9]. 等式中只要有一个数字不一样 就是不一样的 思路: 计算下可以发现, 等式最多只有36个. 然后每个数字i的上界是17-i个 可以预先判掉答案一定是36的, 然后直接暴力搜索每个等式要不要就好了. 注意剪枝即可 ; int a[maxn]; bool flag36; int ans; struct Equation { int x, y…

Rectangles    HDOJ(2056) http://acm.hdu.edu.cn/showproblem.php?pid=2056 题目描述:给2条线段,分别构成2个矩形,求2个矩形相交面积. 算法:先用快速排斥判断2个矩形是否相交.若不相交,面积为0.若相交,将x坐标排序去中间2个值之差,y坐标也一样.最后将2个差相乘得到最后结果. 这题是我大一的时候做过的,当时一看觉得很水,写起来发现其实没我想的那么水.分了好几类情况没做出来.今天看了点关于判断线段相交的知识,想起了这题便拿来练…

Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.   Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines f…

#对coursera上Andrew Ng老师开的机器学习课程的笔记和心得: #注:此笔记是我自己认为本节课里比较重要.难理解或容易忘记的内容并做了些补充,并非是课堂详细笔记和要点: #标记为<补充>的是我自己加的内容而非课堂内容,参考文献列于文末.博主能力有限,若有错误,恳请指正: #---------------------------------------------------------------------------------# 多元线性回归的模型: #-----------…

前言 不说话,先猛戳 Ranklist 看我排名. 这是用 node 自动刷题大概半天的 "战绩",本文就来为大家简单讲解下如何用 node 做一个 "自动AC机". 过程 先来扯扯 oj(online judge).计算机学院的同学应该对 ACM 都不会陌生,ACM 竞赛是拼算法以及数据结构的比赛,而 oj 正是练习 ACM 的 "场地".国内比较有名的 oj 有 poj.zoj 以及 hdoj 等等,这里我选了 hdoj (完全是因为本地上…

Codeforces Round #262 (Div. 2) B B - Little Dima and Equation B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot a…

,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, 1 % Exercise 1: Linear regression with multiple variables %% Initialization %% ================ Part 1: Featu…

题意:有n个点,问其中某一对点的距离最小是多少 分析:分治法解决问题:先按照x坐标排序,求解(left, mid)和(mid+1, right)范围的最小值,然后类似区间合并,分离mid左右的点也求最小值 POJ 3714 #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> const int N = 1e5 + 5; const double INF = 1e…

题目传送门1 2 题意:从无序到有序移动的方案,即最后成1 2 3 4 5 6 7 8 0 分析:八数码经典问题.POJ是一次,HDOJ是多次.因为康托展开还不会,也写不了什么,HDOJ需要从最后的状态逆向搜索,这样才不会超时.判重康托展开,哈希也可. POJ //#include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<string> #include<stack…

Can you solve this equation? Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Now,given the equation *x^ + *x^ + *x^ + *x + == Y,can you find its solution between and ; No…