题面 Accounting for Computer Machinists (ACM) has sufferred from the Y2K bug and lost some vital data for preparing annual report for MS Inc. All what they remember is that MS Inc. posted a surplus or a deficit each month of 1999 and each month when MS Inc.
题目传送门 题意 一家公司，如果盈利，则金额为\(s\),如果亏损，则金额为\(d\),该公司每连续五个月公布一次总财报，已知公布的八次财报全为亏损，问是否盈利，如果盈利则输出最多能盈利的金额，否则输出\(Deficit\)。 思路 这题主要的困难在于理解题目的题意，连续五个月说明是在\(1、234
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 18652 Accepted Submission(s): 7268 Problem Description There are n houses in the village and some bidirectional roads connecting them. Every
Description 给定平面上的 \(n\) 个圆，用三个参数 \((x, y, R)\) 表示圆心坐标和半径。 每次选取最大的一个尚未被删除的圆删除，并同时删除所有与其相切或相交的圆。 最后输出每个圆分别是被那个圆所删除的。 Hint \(1\le n\le 3\times 10^5\) \(0\le |x|, |y|, R \le 10^9\) Solu
There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the ans
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2586 #include<bits/stdc++.h>#define lson rt << 1, l, m#define rson rt << 1 | 1, m + 1, rusing namespace std;typedef long long ll;static const ll inf = (1 << 32);static const int INF
传送门 继续水板子题... #include <bits/stdc++.h>using namespace std;inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch &
传送门 （这次的英文题面要比上一个容易看多了） （英语蒟蒻的卑微） 又是一个很裸的LCT题 （显然，这次不太容易打暴力咧） （但听说还是有大佬用dfs直接a掉了） 正好 趁这个机会复习一下LCT 这里用的是倍增lct的思想（一不小心是会写tle的吧） （也有可能是因为我太弱了才t掉的） 我才不会说我有一
Y2K Accounting Bug Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8282 Accepted: 4106 Description Accounting for Computer Machinists (ACM) has suffe
