#include<bits/stdc++.h> using namespace std; typedef long long LL; ; ; LL a[maxn],b[maxn]; LL quick_pow(LL a,LL b) { LL ans=; while(b) { )ans=ans*a%MOD; a=a*a%MOD; b>>=; } return ans%MOD; } int main() { int T; scanf("%d",&T); whi…

House Man Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3605    Accepted Submission(s): 1517 Problem Description In Fuzhou, there is a crazy super man. He can’t fly, but he could jump from hou…

King Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14791   Accepted: 5226 Description Once, in one kingdom, there was a queen and that queen was expecting a baby. The queen prayed: ``If my child was a son and if only he was a sound kin…

World Exhibition Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2162    Accepted Submission(s): 1063 Problem Description Nowadays, many people want to go to Shanghai to visit the World Exhibiti…

Layout Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15349   Accepted: 7379 Description Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along…

题意: 有N头牛, 有以下关系: (1)A牛与B牛相距不能大于k (2)A牛与B牛相距不能小于k (3)第i+1头牛必须在第i头牛前面 给出若干对关系(1),(2) 求出第N头牛与第一头牛的最长可能距离, 若无解输出-1, 若无限长输出-2 分析: 3个关系对应的 <= 式子是: dis[b] - dis[a] <= d(1) dis[a] - dis[b] <= -d(2) dis[i] - dis[i+1] <= -1(2) 目标式:dis[N] - dis[1] <=…

题意:给a b c要求,b拿的比a拿的多但是不超过c,问你所有人最多差多少 思路:在最短路专题应该能看出来是差分约束,条件是b - a <= c,也就是满足b <= a + c,和spfa的松弛条件相对应,所以我们建一条a到b的边,权值c,然后跑最短路,求出所有差值最大的那个即为答案.应该算是基础的线性差分约束题. ps:队列超时,这里用栈. 关于差分约束可以点这里 #include<cstdio> #include<set> #include<map> #…

House Man Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2056    Accepted Submission(s): 811 Problem Description In Fuzhou, there is a crazy super man. He can’t fly, but he could jump from hou…

可爱的菜菜子 题目链接:https://cometoj.com/contest/38/problem/D?problem_id=1543 数据范围:略. 题解: 首先,如果第一个操作是单点修改,我们就只需要用线段树维护区间线性基即可. 但是,第一个操作是区间异或,这就很难受了. 区间操作可以怎么转化成单点?差分呗. 将原序列异或差分之后,确实修改变成了单点修改. 查询是否有影响?其实是没有影响的因为相当于整个一个区间的每个数异或上同一个数(前$l - 1$个数的异或和),线性基不受影响. 代码:…

题意:给你一个数组,有两种操作,一种区间xor一个值,一个是查询区间xor的结果的种类数 做法一:对于一个给定的区间,我们可以通过求解线性基的方式求出结果的种类数,而现在只不过将其放在线树上维护区间线性基. 用线段树去维护区间合并 #include <bits/stdc++.h> using namespace std; const int maxn = 1e5; struct node { ],lazy,st; void init() { memset(bas,,sizeof(bas));…