题目描述 Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1 might h…

题意:给你一组数,询问\(q\)次,问所给区间内的最大值和最小值的差. 题解:经典RMQ问题,用st表维护两个数组分别记录最大值和最小值然后直接查询输出就好了 代码: int n,q; int a[N]; int dp1[N][30],dp2[N][30]; int lg[N]; void lg_Init(){ for(int i=1;i<=n;++i){ int k=0; while(1<<(k+1)<=i) k++; lg[i]=k; } } void RMQ_Init1(){…

维护区间最值的模板题. 1.树状数组 1 #include<bits/stdc++.h> 2 //树状数组做法 3 using namespace std; 4 const int N=5e4+10; 5 int m,ma[N],mi[N],n,c[N]; 6 7 int lowbit(int x){ 8 return x&(-x); 9 } 10 11 void ins(int x,int v){ 12 while(x<=n){ 13 ma[x]=max(ma[x],v);mi…

Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10924 Accepted: 3244 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by h…

Gold Balanced Lineup Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13540   Accepted: 3941 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared…

P1360 [USACO07MAR]Gold Balanced Lineup G (前缀和+思维) 前言 题目链接 本题作为一道Stl练习题来说,还是非常不错的,解决的思维比较巧妙 算是一道不错的题 思路分析 第一眼看到这题,我还以为是数据结构题,看来半天没看出来数据结构咋做(我还是太菜了) 我们对\(m\)种能力有\(n\)次操作,需要找到对每种能力提升相同的最大操作区间的长度,求最大 区间,我们考虑维护这\(m\)种技能提升值的前缀和,假设第\(l+1\)次操作到第\(r\)次操作对\(m\…

Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 13215 Accepted: 3873 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by h…

1702: [Usaco2007 Mar]Gold Balanced Lineup 平衡的队列 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 510  Solved: 196[Submit][Status][Discuss] Description Farmer John's N cows (1 <= N <= 100,000) share many similarities. In fact, FJ has been able to narrow…

约数和 题目描述 给出a和b求a^b的约数和. 输入格式: 一行两个数a,b. 输出格式: 一个数表示结果对 9901 的模. Input 2 3 Output 15 SB的思路: 这是一道典型的数论题,本蒟蒻在做的时候首先瞄出a为质数的解法(简直废话,是个人都看得出), 即sum(a,b)=a^0+a^2+a^3+···+a^(b-1)+a^b,然后自以为搞出了什么,结果随手举个反例就Wa了,但是很明显也很容易想到要用快速幂. 然后我又想到洛谷月赛T1,以及一道要用到费马小定理的题目,加上我打…

题目:http://poj.org/problem?id=3274 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip> #include<cmath> #include<map> #include<ve…