题目地址:http://poj.org/problem?id=1064 有N条绳子,它们的长度分别为Ai,如果从它们中切割出K条长度相同的绳子,这K条绳子每条最长能有多长. 二分绳子长度,然后验证即可.复杂度o(nlogm) #include<cstdio> #include<iostream> #include<string.h> #include<algorithm> #include<math.h> #include<stdbool.…

Cable master Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21071   Accepted: 4542 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to…

Cable master Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21071   Accepted: 4542 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to…

题目链接:http://poj.org/problem?id=1064 有n条绳子,长度分别是Li.问你要是从中切出m条长度相同的绳子,问你这m条绳子每条最长是多少. 二分答案,尤其注意精度问题.我觉得关于浮点数的二分for循环比while循环更好一点.注意最后要用到floor 保证最后答案不会四舍五入. #include <iostream> #include <cstdio> #include <cmath> using namespace std; int n ,…

地址 http://poj.org/problem?id=1064 题解 二分即可 其实 对于输入与精度计算不是很在行 老是被卡精度 后来学习了一个函数 floor 向负无穷取整 才能ac 代码如下 #include <iostream> #include <vector> #include <math.h> #include <algorithm> using namespace std; vector<double> v; int n, k;…

题目: 给n个长度为l[i](浮点数)的绳子,要分成k份相同长度的 问最多多长 题解: 二分长度,控制循环次数来控制精度,输出也要控制精度<wa了好多次> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #define N 10010 using namespace std; double L[N],l,r,mid; long long n,k; //y…

题意:给定 n 条绳子,它们的长度分别为 ai,现在要从这些绳子中切出 m 条长度相同的绳子,求最长是多少. 析:其中就是一个二分的水题,但是有一个坑,那么就是最后输出不能四舍五入,只能向下取整. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include &l…

POJ 1064 Cable master 一开始把 int C(double x) 里面写成了  int C(int x) ,莫名奇妙竟然过了样例,交了以后直接就wa. 后来发现又把二分查找的判断条件写错了,wa了n次,当 c(mid)<＝k时,令ub＝mid,这个判断是错的,因为要找到最大切割长度,当满足这个条件时,可能已经不是最大长度了,此时还继续缩小区间,自然就wa了,(从大到小递减,第一次满足这个条件的值,就是最大的值),正确的判断是当 c(mid)<k时,令ub＝mid,这样循环1…

poj 1064 Cable master 判断一个解是否可行 浮点数二分 题目链接: http://poj.org/problem?id=1064 思路: 二分答案,floor函数防止四舍五入 代码: #include <iostream> #include <stdio.h> #include <math.h> #include <algorithm> using namespace std; const int maxn = 10005; const…

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…

题目链接 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to organize the most honest contest ever. It was decided to connect computers for the contestan…

Cable master Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 29554   Accepted: 6247 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to…

题目链接:http://poj.org/problem?id=1064 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to organize the most honest contest ever. It was decided to conn…

题目链接: 传送门 Cable master Time Limit: 1000MS     Memory Limit: 65536K 题目描述 有N条绳子,它们长度分别为Li.如果从它们中切割出K条长度相同的绳子的话,这K条绳子每条最长能有多长?答案保留小数点后两位. 思路 二分搜索答案 #include<iostream> #include<cstdio> #include<cmath> #define EPS 1e-6 const int INF = 100000;…

Cable master Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 37865   Accepted: 8051 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to…

Cable master Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 65358   Accepted: 13453 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to…

http://poj.org/problem?id=1064 题目大意: 有N条绳子,他们的长度分别为Li,如果从它们中切割出k条长度相同的绳子的话,这K条绳子每条能有多长? 思路: 二分,设答案为mid=(L+R)/2, 如果以mid划分可以分割出不小于K条绳子,那么解>=mid, 否则解小于mid PS: 最后的输出坑死了.要保留两位整数且不进位....T T #include<cstdio> #include<cmath> const int MAXN= 10000+1…

嗯... 题目链接:http://poj.org/problem?id=1064 其实这是一道很好想的二分答案的一道题... 二分的区间就是1~max_l,从1开始是因为所有小于1的都需要按0计算,没必要讨论了... 每次二分出来的答案看它能否把电缆切成大于等于k块,如果可以,我们不能保证它是最优的答案,所以要向更大的地方二分:如果现在都不可以,我们必须向更小的地方二分,才有可能可以. 这道题注意二分一般在整数中二分,所以我们先把它们都乘100,如果要求精度更高,则乘的数更大,然后再整数二分,最…

给出n根绳子,求把它们分割成K条等长的绳子的最大长度是多少? 二分 用 for(int i=0; i<100; ++i) 取代   while(r-l>eps) 循环100次精度能达到1e-30,基本上能一般题目的精度要求. 而 浮点数二分区间的话easy产生精度缺失导致死循环. #include<cstdio> double L[10000 + 10]; int n, k; int ok(double x) { int cnt = 0; for(int i=0; i<n;…

题目: 这题有点坑,G++过不了,C++能过. 条件:n个数据a[],分成k段,结果精度要求两位小数. 问题:每段最长为多少? 思路:因为精度要求为两位小数,我先把所有的长度a[]*100. 我们对答案二分搜索,把l设置为0,r设置为1000*10000*100+1(数据量*每个数据最大的大小*精度+1). 这样我们搜索的数就不用处理精度了,我们可以二分算出结果然后除以100. 代码: #include <iostream> #include <algorithm> #includ…

和杭电那一题一样,只不过G++交不能通过,C++能过 wa了好多好多好多次----------------------------------------- #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #include<vector> #include<map> #include<set&…

POJ 3273 Monthly Expense二分查找(最大值最小化问题) 题目:Monthly Expense Description Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ mon…

POJ 3273 Monthly Expense 此题与POJ3258有点类似,一开始把判断条件写错了,wa了两次,二分查找可以有以下两种: ){ mid=(lb+ub)/; if(C(mid)<=m) ub=mid; ; //此时下限过小 } out(ub);//out(lb) 我一开始是写的下面这种,下面这种要单独判断lb和ub的值,因为用下面这种判断lb,ub都可能成立 ){ mid=(lb+ub)/; if(C(mid)<=m) ub=mid; else lb=mid; } if(C(…

Cable master 求电缆的最大长度(二分法)   Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to organize the most honest contest ever. It was decided to connect com…

题目地址:http://poj.org/problem?id=2456 最大化最小值问题.二分牛之间的间距,然后验证. #include<cstdio> #include<iostream> #include<string.h> #include<algorithm> #include<math.h> #include<stdbool.h> #include<time.h> #include<stdlib.h>…

POJ 3579 题意 双重二分搜索:对列数X计算∣Xi – Xj∣组成新数列的中位数 思路 对X排序后,与X_i的差大于mid(也就是某个数大于X_i + mid)的那些数的个数如果小于N / 2的话,说明mid太大了.以此为条件进行第一重二分搜索,第二重二分搜索是对X的搜索,直接用lower_bound实现. #include <iostream> #include <algorithm> #include <cstdio> #include <cmath&g…

题目地址:http://poj.org/problem?id=3122 二分每块饼的体积.为了保证精度,可以先二分半径的平方r*r,最后再乘以PI.要注意一点,要分的人数要包括自己,及f+1. #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> using namespace std; ; const double PI…

题目描述 Given a list of N integers with absolute values no larger than 10 15, find a non empty subset of these numbers which minimizes the absolute value of the sum of its elements. In case there are multiple subsets, choose the one with fewer elements.…

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #define N 2000 #define M 1000010 #define inf 1<<30 using namespace std; struct Edge{ int to,val,next; }edge[M]; ]; void addedge(int from,int to,int val…

题目链接:点击打开链接 有n段绳子,给定n段绳子的长度,单位为厘米.求能够把这些绳子分成k段的最长的段的长度.题目中的trick是最小是1cm,长度不能小于1cm,因此要转换成int来解,然后二分可以截得的绳子的长度. #include <iostream> #include <string> #include <algorithm> using namespace std; #define INF 100000000 int l[10009]; int n, k; b…