[POJ] 1064 Cable master (二分查找)

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)

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<string.h>
 4 #include<algorithm>
 5 #include<math.h>
 6 #include<stdbool.h>
 7 #include<time.h>
 8 #include<stdlib.h>
 9 #include<map>
10 #include<stack>
11 #include<queue>
12 #include<vector>
13 using namespace std;
14 #define clr(x,y)    memset(x,y,sizeof(x))
15 #define sqr(x)      ((x)*(x))
16 #define rep(i,a,b)  for(int i=(a);i<=(b);i++)
17 #define LL          long long
18 #define INF         0x3f3f3f3f
19 #define A           first
20 #define B           second
21 const int N=10000+131;
22 int     n,k;
23 double  a[N];
24
25 bool check(double len)
26 {
27     int num=0;
28     for(int i=0;i<n;i++) {
29         num+=(int)(a[i]/len);
30     }
31     return num>=k;
32 }
33
34
35 void solve()
36 {
37     double lb=0,ub=INF;
38
39     scanf("%d%d",&n,&k);
40     for(int i=0;i<n;i++) {
41         scanf("%lf",&a[i]);
42     }
43
44     for(int i=0;i<100;i++) {
45         double mid=(lb+ub)/2;
46         if(check(mid)) {
47             lb=mid;
48         } else {
49             ub=mid;
50         }
51     }
52      printf("%.2f\n",floor(ub*100)/100);
53 }
54
55 int main()
56 {
57     solve();
58
59     return 0;
60 }
时间: 2024-10-27 06:19:19

[POJ] 1064 Cable master (二分查找)的相关文章

POJ 1064 Cable master (二分 分数化整数)

Cable master Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28764   Accepted: 6091 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 (二分答案)

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. 1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 using namespace std;

POJ 1064 Cable master (二分)

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(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

POJ 1064 Cable master 浮点数二分

Cable master Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21181   Accepted: 4571 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 精度!!!

给出n根绳子,求把它们切割成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) 代替   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容易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; ++

二分搜索 POJ 1064 Cable master

题目传送门 1 /* 2 题意:n条绳子问切割k条长度相等的最长长度 3 二分搜索:搜索长度,判断能否有k条长度相等的绳子 4 */ 5 #include <cstdio> 6 #include <algorithm> 7 #include <cstring> 8 #include <cmath> 9 using namespace std; 10 11 const int MAXN = 1e4 + 10; 12 const int INF = 0x3f3f

[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

poj 1064 Cable master【浮点型二分查找】

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