Magician Problem Description Fantasy magicians usually gain their ability through one of three usual methods: possessing it as an innate talent, gaining it through study and practice, or receiving it from another being, often a god, spirit, or demon…

传送门 给出一个数字字串,给出若干个询问,询问在字串的一段区间保证出现2017" role="presentation" style="position: relative;">20172017而不出现2016" role="presentation" style="position: relative;">20162016需要删去的最少字符. 下面定义5个状态. 状态0:连2都没有 状态1:只…

小明系列问题--小明序列 Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Problem Description 大家都知道小明最喜欢研究跟序列有关的问题了,可是也就因为这样,小明几乎已经玩遍各种序列问题了.可怜的小明苦苦地在各大网站上寻找着新的序列问题,可是找来找去都是自己早已研究过的序列.小明想既然找不到,那就自己来发明一个新的序列问题吧!小明想啊想,终于想出了一个新的序列问题,…

Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Problem Description The inversion number of a given number sequence a1, a2, -, an is the number of pairs (ai, aj) that satisfy i < j and ai > a…

Color it Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others) Problem Description Do you like painting? Little D doesn't like painting, especially messy color paintings. Now Little B is painting. To prevent him from dr…

Fantasy magicians usually gain their ability through one of three usual methods: possessing it as an innate talent, gaining it through study and practice, or receiving it from another being, often a god, spirit, or demon of some sort. Some wizards ar…

题目链接:Magician 题意: 给你一个长度为n的序列v,你需要对这个序列进行m次操作,操作一共有两种,输入格式为 type a b 1.如果type==0,你就需要输出[a,b]区间内的美丽序列中所有元素的和,要使得这个值尽可能大 2.如果type==1,你就需要把a位置的元素值改为b 区间[a,b]的美丽序列就是va,va+1...vb.你需要从中取出任意个元素,这些元素的位置必须是奇偶交替 例如给你一个序列1,2,3,4,5,6,7 你取出来的美丽序列就有可能使1,2,3,4,5,6,…

题目中没有明说会爆int和longlong 的精度,但是在RNG函数中不用unsigned int 会报精度,导致队友debug了很久... 根据每次生成的l,r,v对区间更新m次,然后求 i*ai的异或和.挺裸的线段树,只要ai<v就更新ai.在线段树结点中维护区间最小值,每次更新时,如果区间最小值>=v,那么就不用更新了. #include<bits/stdc++.h> #define lson rt << 1 #define rson rt << 1…

设读入的数组是a,树状数组用来维护a数组区间和sum,线段树用来维护一个另一个数组ssum的区间和,区间每个点a[i]*(n-i+1),那么l-r的答案是l-r的ssum-(n-r)*(sum[r]-sum[l-1]) (纸上画一下就知道了) #include<bits/stdc++.h> using namespace std; const int maxn=100010; int n,q,ql,qr,p; long long v,_sum; long long sum[maxn],ssum…

线段树,假设求(x1,y1)点的贡献,就找所有比该点出现时间晚,且x坐标大于x1的点中y最大的,贡献就是Y-y1,由于题目条件限制,不可能有x坐标大于(x1,y1)且y坐标大于y1的点,所以贡献肯定为正. 思路参考了这篇博客:https://blog.csdn.net/qq_39599067/article/details/82560005#accode. #include<bits/stdc++.h> using namespace std; ; struct node{ int x,y;…