【POJ - 3579 】Median（二分）

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POJ 3579 Median 二分加判断

Median Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12453 Accepted: 4357 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < j ≤ N). We can get C(N,2) differenc…

POJ 3579 Median (二分）

Median Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7423 Accepted: 2538 Description Given N numbers, X1, X2, ... , XN, let us calcu…

poj 3579 Median 二分套二分或二分加尺取

Median Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5118 Accepted: 1641 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < j ≤ N). We can get C(N,2)differences…

POJ 3579 Median（二分答案+Two pointers）

[题目链接] http://poj.org/problem?id=3579 [题目大意] 给出一个数列,求两两差值绝对值的中位数. [题解] 因为如果直接计算中位数的话,数量过于庞大,难以有效计算, 所以考虑二分答案,对于假定的数据,判断是否能成为中位数此外还要使得答案尽可能小,因为最小的满足是中位数的答案,才会是原差值数列中出现过的数对于判定是不是差值的中位数的过程,我们用尺取法实现. 对于差值类的题目,还应注意考虑边界,即数列只有一位数的情况. [代码] #include <cstdio…

POJ 3579 Median（二分答案）

Median Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11599 Accepted: 4112 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < j ≤ N). We can get C(N,2) differences t…

POJ 3579 Median 【二分答案】

<题目链接> 题目大意: 给出 N个数,对于存有每两个数的差值的序列求中位数,如果这个序列长度为偶数个元素,就取中间偏小的作为中位数. 解题分析: 由于本题n达到了1e5,所以将这些数之间的差值全部求出来显然是不可行的,这里用的是二分答案.先通过二分,假设枚举出的答案为mid,即,这些数字差值绝对值的中位数为mid.然后我们在通过二分查找,对每一个数字,查找它后面的所有满足与它的差值小于等于mid的数的个数,即查找所有差值小于等于mid的对数.因为中位数所在的编号很容易求得,为 (n*(n-1…

poj 3579 Median （二分搜索之查找第k大的值）

Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ ( ≤ i < j ≤ N). We can ) differences through this work, and now your task is to find the median of the differences as quickly as you ca…