题目链接:http://poj.org/problem?id=3579 Median Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8286   Accepted: 2892 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < …

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 through this work, and now your task is to find the median of the differences as quickly…

Median Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5468   Accepted: 1762 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) difference…

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…

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

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…

C. Maximum Median 题意: 给定一个数组,可每次可以选择一个数加1,共执行k次,问执行k次操作之后这个数组的中位数最大是多少? 题解:首先对n个数进行排序,我们只对大于中位数a[n/2]的数进行操作,所以这个最大中位数的取值范围是确定的,在区间[  [a[n/2],a[n-1]  ]之内,二分枚举最大的中位数x; 通过判断使x成为最大中位数的操作数是否大于k来缩小范围 #include<iostream> #include<string.h> #include<…

Median http://poj.org/problem?id=3579 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11225   Accepted: 4016 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < j ≤ …

Median Descriptions 给N数字, X1, X2, ... , XN,我们计算每对数字之间的差值:∣Xi - Xj∣ (1 ≤ i < j ≤N). 我们能得到 C(N,2) 个差值,现在我们想得到这些差值之间的中位数. 如果一共有m个差值且m是偶数,那么我们规定中位数是第(m/2)小的差值. Input 输入包含多测每个测试点中,第一行有一个NThen N 表示数字的数量.接下来一行由N个数字:X1, X2, ... , XN( Xi ≤ 1,000,000,000  3 ≤…

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…