【差分】luogu_P4231 三步必杀

题意

对一个数组的某个区间加上一段等差数列，求出若干个操作后每个元素的值。

思路

考虑差分。
设$A$为原数组，$B$为一阶差分数组，$C$为二阶差分数组。
对$[l,r]$加上首项为$s$，末项为$e$，公差为$d$的等差数列，可得
$A_i=A_i+s+(i-l)\ast d$

$B_l=(A_l+s)-A_{l-1}=B_l+s$
$B_i=(A_i+s\ast (i-l)\ast d)-(A_{i-1}+s\ast (i-1-l)\ast d)=B_i+d$
$B_{r+1}=A_{r+1}-(A_r+e)=B_{r+1}-e$

$C_l=(B_l+s)-(B_{l-1})=C_l+s$
$C_{l+1}=(B_{l+1}+d)-(B_l+s)=C_l+d-s$
$C_{r+1}=(B_{r+1}-e)-(B_r+d)=C_l-d-e$
$C_{r+2}=B_{r+2}-(B_{r+1}-e)=C_{r+2}+e$

故只用修改4个值，做两次前缀和即可。

代码

#include <cstdio>
#include <algorithm>
#define int long long

const int N = 1e7 + 5;
int n, m, ans1, ans2;
int a[N];

signed main() {
	scanf("%lld %lld", &n, &m);
	for (int i = 1, l, r, s, e; i <= m; i++) {
		scanf("%lld %lld %lld %lld", &l, &r, &s, &e);
		int d = (e - s) / (r - l);
		a[l] += s;
		a[l + 1] += d - s;
		a[r + 1] -= d + e;
		a[r + 2] += e;
	}
	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 = std::max(ans2, a[i]);
	printf("%lld %lld", ans1, ans2);
}