题意：

在这里插入图片描述

思路：

考虑给 $[2, 6]$ 加上 $s = 2, e = 10$ 的等差数列，变成 $2, 4, 6, 8, 10$ ，考虑差分数组 $2, 2, 2, 2, 2, - 10$ ，即 $a [l] + = a, a [l + 1 - > r] + = d, a [r + 1] - = e$ ，这个显然可以用树状数组维护，但是我们可以继续对差分数组差分，让 $a [l] + = a, a [l + 1] + = d - a, a [r + 1] + = - d - e, a [r + 2] + = e$ 。
这样最后求两次前缀和，让后输出答案即可。

// Problem: P4231 三步必杀
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P4231
// Memory Limit: 250 MB
// Time Limit: 500 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=10000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,m;
LL a[N];int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);	cin>>n>>m;while(m--) {LL l,r,s,e; scanf("%lld%lld%lld%lld",&l,&r,&s,&e);LL d=(e-s)/(r-l);a[l]+=s; a[l+1]+=d-s; a[r+1]+=-d-e; a[r+2]+=e;}LL ans1=0,ans2=0;for(int i=1;i<=n;i++) a[i]+=a[i-1];for(int i=1;i<=n;i++) {a[i]+=a[i-1];ans1^=a[i]; ans2=max(ans2,a[i]);}printf("%lld %lld\n",ans1,ans2);return 0;
}
/**/