poj 1064 Cable master 判断一个解是否可行浮点数二分

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poj 1064 Cable master 判断一个解是否可行浮点数二分

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…

POJ 1064 Cable master（二分查找+精度）（神坑题）

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…

二分搜索 POJ 1064 Cable master

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…

Poj：1064 ：：Cable master (假定一个解并判断是否可行）（二分搜索答案）

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 contestants us…

[ACM] poj 1064 Cable master （二分查找）

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…

[ACM] poj 1064 Cable master （二进制搜索）

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…

POJ 1064 Cable master （二分查找）

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…