Interviewe Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6689    Accepted Submission(s): 1582 Problem Description YaoYao has a company and he wants to employ m people recently. Since his compa…

Interviewe Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6722    Accepted Submission(s): 1592 Problem Description YaoYao has a company and he wants to employ m people recently. Since his compa…

人生中第一次写RMQ....一看就知道 RMQ+2分但是题目文不对题....不知道到底在问什么东西....各种WA,TLE,,RE...后就过了果然无论错成什么样都可以过的,就是 上层的样例 啊  Interviewe Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3865    Accepted Submission(s): 956…

Interviewe Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 945 Accepted Submission(s): 234   Problem Description YaoYao has a company and he wants to employ m people recently. Since his company is…

Interviewe Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4543    Accepted Submission(s): 1108 Problem Description YaoYao has a company and he wants to employ m people recently. Since his compa…

题目大意:给定n个数的序列,让我们找前面k个区间的最大值之和,每个区间长度为n/k,如果有剩余的区间长度不足n/k则无视之.现在让我们找最小的k使得和严格大于m. 题解:二分k,然后求RMQ检验. ST算法: #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int maxn=200010; int d[maxn][30]; int a[maxn];…

还是挺简单的,但是区间处理的时候要注意一下 #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<cmath> using namespace std; #define maxn 200005 #define ll long long ],a[maxn],n; ll k; void ST(){ memset(dpmax,,sizeof…

面试n个人,可以分任意组数,每组选一个,得分总和严格大于k,问最少分几组 就是暴力嘛...想到就去写吧.. #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <…

题意: 将\(n\)个数分成\(m\)段相邻区间,每段区间的长度为\(\left \lfloor \frac{n}{m} \right \rfloor\),从每段区间选一个最大值,要让所有的最大值之和大于\(k\).求最小的\(m\). 分析: 预处理RMQ,维护区间最大值. 然后二分\(m\),将每段区间最大值加起来判断即可. #include <cstdio> #include <cstring> #include <algorithm> using namespa…

#include <cstdio> #include <iostream> #include <algorithm> using namespace std; + ; int a[maxn]; ]; int n, k; void RMQ(int num){ ; i <= num; i++){ maxsum[i][] = a[i]; } ; j < ; j++){ ; i <= num; i++){ << j) - <= num){ m…