P4231 三步必杀（二阶差分）

题目

输入输出样例
输入样例一

5 2
1 5 2 10
2 4 1 1

输出样例一

3 10

输入样例二

6 2
1 5 2 10
2 4 1 1

输出样例二

3 10

题解

• 设$a[]$为原数组，$b[]$为$a[]$的差分数组，$c[]$为$b[]$的差分数组
• 对于$l,r,s,t$我们可以算出公差为$$d=\frac{t-s}{r-l}$$
• 那么$a[]$数组中数字的改变为$$a[i]=a[i]+s+d\ast (i-l)(l\le i\le r)$$
• $b[]$数组中数字($b[i]=a[i]-a[i-1]$)的改变为
  $$b[l]=a[l]+s-a[l-1]=b[l]+s$$
  $$b[i]=a[i]+d\ast (i-l)-a[i-1]-d\ast (i-1-l)=b[i]+d(l+1\le i\le r)$$
  $$b[r+1]=a[r+1]-(a[r]+t)=b[r+1]-t$$
• $c[]$数组中数字($c[i]=b[i]-b[i-1]$)的改变为
  $$c[l]=b[l]+s-b[l-1]=c[l]+s$$
  $$c[l+1]=b[l+1]+d-(b[l]+s)=c[l+1]+d-s$$
  $$c[i]=(b[i]+d)-(b[i-1]+d)=c[i](l+2\le i\le r)$$
  $$c[r+1]=(b[r+1]-t)-(b[r]+d)=c[r+1]-t-d$$
  $$c[r+2]=b[r+2]-(b[r+1]-t)=c[r+2]+t$$
• 统计时求两遍前缀和即可

code

#include  
using namespace std; 
typedef long long LL; 
const int maxn = 1e7 + 100; 

template <typename T> 
inline void read(T &s) {
	s = 0; 
	T w = 1, ch = getchar(); 
	while (!isdigit(ch)) { if (ch == '-') w = -1; ch = getchar(); }
	while (isdigit(ch)) { s = (s<<1) + (s<<3) + (ch^48); ch = getchar(); }
	s *= w; 
}

int n, m; 
LL a[maxn]; 

int main() {
	read(n), read(m); 
	for (int i = 1; i <= m; ++i) {
		LL l, r, s, t; 
		read(l), read(r), read(s), read(t);  
		LL d = (t-s) / (r-l); // 公差 
		a[l] += + s; 
		a[l+1] += d - s; 
		a[r+1] -= d + t; 
		a[r+2] += t; 
	}
	LL sum1 = 0ll, sum2 = 0ll, ans = 0ll, Max = 0ll;
	for (int i = 1; i <= n; ++i) {
		sum1 += a[i]; 
		sum2 += sum1; 
		Max = max(Max, sum2); 
		ans ^= sum2; 
	}
	printf("%lld %lld\n", ans, Max); 
	return 0; 
}