[[https://codeforces.com/gym/102569|比赛链接]]

====== 题解 ======

===== A. Array's Hash =====

=== 题意 ===

给定一个长度为$n$的数组，这么定义该数组的哈希值：每次从数组开头取出两个数，将后一个数减去前一个数得到的数值放入数组开头，如此重复，直到数组中只剩下一个数，最后这个数便为数组的哈希值。现在有$m$次操作，每次操作把一段区间的数加上$v$，要求输出每次操作后数组的哈希值

===题解===

显然数组的哈希值为$\sum_{i=1}^n{\left(-1\right)^\left(n-i\right)\times a_i}$

因此当区间左右端点奇偶性相同时对哈希值无贡献，奇偶性不同时如果区间左端点与$n$奇偶性相同则哈希值加$v$，否则减$v$

时间复杂度$O\left(n+m\right)$

===== B. Bonuses on a Line =====

=== 题意 ===

数轴上有 $n$ 份奖金，每份奖金的坐标为 $x_i$ ，总共有 $t$ 秒的时间，每秒可走 $1$ 的距离。初始在原点 $0$ 位置，问最多能获得多少份奖金？

=== 题解 ===

先向某一方向跑，然后再折返跑向另一方向。例如先向负方向跑，在每份奖金处，利用二分查找，找到能够折返跑到正方向的奖金的最大份数即可。

时间复杂度 $O\left(n\log n\right)$

=== 代码 ===

<code cpp>

#include<bits/stdc++.h>
using namespace std;
const int N=200005;
typedef long long ll;
vector<ll>neg;
vector<ll>pos;
int main(){
	ll n,t;
	scanf("%lld %lld",&n,&t);
	for(ll i=1;i<=n;i++){
		ll x;
		scanf("%lld",&x);
		if(x<0) neg.push_back(-x);
		else pos.push_back(x);
	}
	reverse(neg.begin(),neg.end());
	neg.push_back(1e15);pos.push_back(1e15);
	ll maxx=0;
	for(ll i=0;i<neg.size()-1;i++){
		if(t>=neg[i]) maxx=max(maxx,i+1);
		ll left=t-2*neg[i];
		ll p=upper_bound(pos.begin(),pos.end(),left)-pos.begin()-1;
		if(p>=0) maxx=max(maxx,i+1+p+1);
	}
	for(ll i=0;i<pos.size()-1;i++){
		if(t>=pos[i]) maxx=max(maxx,i+1);
		ll left=t-2*pos[i];
		ll p=upper_bound(neg.begin(),neg.end(),left)-neg.begin()-1;
		if(p>=0) maxx=max(maxx,i+1+p+1);
	}
	printf("%lld",maxx);
	return 0;
}

</code>

===== C. Manhattan Distance =====

===题意===

在直角坐标系中给定$n$整点$\left(-10^8\le x_i,y_i\le 10^8\right)$，可以得到$\frac{n\times \left(n-1\right)}{2}$个点对，将所有点对的哈密顿距离排序，要求输出第k大的哈密顿距离$\left(2\le n\le 100000,1\le k\le \frac{n\times \left(n-1\right)}{2}\right)$

===题解===

大概思路为二分答案$d$，统计哈密顿距离$\le d$的点对个数

首先，将坐标系顺时针旋转$45$度，放大$\sqrt{2}$倍，所以所有点坐标变为$\left(x-y,x+y\right)$，与某个点哈密顿距离$\le d$转化为在以该点为中心的边长为$2d$的网格正方形中

考虑用滑动窗口+树状树组统计答案，具体过程见代码

时间复杂度$O\left(\log\left(4\times 10^8\right)  n\log n\right)$

===代码===

<code cpp>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cctype>
#define _for(i,a,b) for(int i=(a);i<(b);++i)
#define _rep(i,a,b) for(int i=(a);i<=(b);++i)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
inline int read_int(){
	int t=0;bool sign=false;char c=getchar();
	while(!isdigit(c)){sign|=c=='-';c=getchar();}
	while(isdigit(c)){t=(t<<1)+(t<<3)+(c&15);c=getchar();}
	return sign?-t:t;
}
inline LL read_LL(){
	LL t=0;bool sign=false;char c=getchar();
	while(!isdigit(c)){sign|=c=='-';c=getchar();}
	while(isdigit(c)){t=(t<<1)+(t<<3)+(c&15);c=getchar();}
	return sign?-t:t;
}
#define lowbit(x) (x)&(-x)
const int MAXN=1e5+5;
struct Node{
	int x,y;
	bool operator < (const Node &b)const{
		return x<b.x||(x==b.x&&y<b.y);
	}
}node[MAXN];
int n,m,c[MAXN],Y[MAXN];
LL k;
void add(int pos,int v){
	while(pos<=n){
		c[pos]+=v;
		pos+=lowbit(pos);
	}
}
int query(int pos){
	int ans=0;
	while(pos){
		ans+=c[pos];
		pos-=lowbit(pos);
	}
	return ans;
}
LL Count(int d){
	mem(c,0);
	LL ans=0;
	for(int i=1,j=1;i<=n;i++){
		while(j<i&&node[i].x-node[j].x>d){
			int pos=lower_bound(Y+1,Y+m,node[j].y)-Y;
			add(pos,-1);
			j++;
		}
		int pos1=upper_bound(Y+1,Y+m,node[i].y+d)-Y-1;
		int pos2=lower_bound(Y+1,Y+m,node[i].y-d)-Y-1;
		int pos3=lower_bound(Y+1,Y+m,node[i].y)-Y;
		ans+=query(pos1)-query(pos2);
		add(pos3,1);
	}
	return ans;
}
int main()
{
	n=read_int(),k=read_LL();
	int x,y;
	_rep(i,1,n){
		x=read_int(),y=read_int();
		node[i].x=x-y,node[i].y=x+y;
		Y[i]=x+y;
	}
	sort(node+1,node+n+1);
	sort(Y+1,Y+n+1);
	m=unique(Y+1,Y+n+1)-Y;
	int lef=1,rig=4e8,mid,ans=-1;
	while(lef<=rig){
		mid=lef+rig>>1;
		if(Count(mid)<k)
		lef=mid+1;
		else{
			ans=mid;
			rig=mid-1;
		}
	}
	cout<<ans;
	return 0;
}
</code>

===== D. Lexicographically Minimal Shortest Path =====

===== E. Fluctuations of Mana =====

签到题
===== F. Moving Target =====

===== G. Nuts and Bolts =====

===== H. Tree Painting =====

===== I. Sorting Colored Array =====

===== J. The Battle of Mages =====

===== K. Table =====

===== L. The Dragon Land =====

===== M. Notifications =====

签到题

====== 总结 ======