Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3448 Accepted Submission(s): 1144 Problem Description It has been ten years since TJU-ACM established. And in this year all the retired TJU-ACM
Meeting point-1 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2866 Accepted Submission(s): 919 Problem Description It has been ten years since TJU-ACM established. And in this year all the
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4311 解题报告:在一个平面上有 n 个点,求一个点到其它的 n 个点的距离之和最小是多少. 首先不得不说一下做这道题囧的事,杭电用的是__int64,我前面定义的时候用的是__int64,然后后面输出结果的时候格式控制符竟然用了%lld,还小卡了一会,唉,大意了啊. 然后感觉这题好巧妙,做法是将 x 与 y的距离分开求,我也是看了学长博客之后才懂的http://www.cnblogs.com/Lyu
求中位数,注意求中位数前排序.... #include <bits/stdc++.h> using namespace std; #define LL long long const int MAXN=1e4+10; LL a[MAXN]; int main() { int n; cin>>n; LL sum=0; for(int i=0;i<n;i++) { cin>>a[i]; } sort(a,a+n); int mid=(n+1)/2-1; // cout
题目链接 正经解法: 给定n个点的坐标,找一个点,到其他点的曼哈顿距离之和最小.n可以是100000.大概要一个O(nlogn)的算法.算曼哈顿距离可以把x和y分开计算排好序后计算前缀和就可以在O(1)时间内判断一个点到其他点的距离. #include<cstdio> #include<algorithm> using namespace std; #define ll long long #define N 100005 int t,n; ll ans,sum[N],sx[N],