正题

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题目大意

修改$[L..R]$加上$[S..E]$的等差数列，求最终答案。

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题目大意

很明显的差分。

$a_i$为原数组，$b_i$为一阶差分数组，$c_i$为二阶差分数组

$$a_x=a_x+s+(x-l)\ast k(x\in [l..r])$$
然后
$$b_l=b_l+s$$
$$b_x=b_x+d(x\in [l+1..r])$$
$$b_{r+1}=b_{r+1}-t$$
$$c_l=c_l+s$$
$$c_{l+1}=c_l+d-s$$
$$c_r=c_l-d-t$$
$$c_{r+1}=c_{r+2}+t$$
其实弄3个数组搞一搞也行

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$code$

#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=1e7+10;
ll n,m,f[N],z,maxs;
int main()
{scanf("%lld%lld",&n,&m);for(ll i=1;i<=m;i++){ll l,r,s,e,k;scanf("%lld%lld%lld%lld",&l,&r,&s,&e);k=(e-s)/(r-l);f[l]+=s;f[l+1]+=k-s;f[r+1]-=k+e;f[r+2]+=e;}ll x=0,y=0;for(ll i=1;i<=n;i++){x=x+f[i];y+=x;z^=y;maxs=max(maxs,y);}printf("%lld %lld",z,maxs);
}