[luoguP2221] [HAOI2012]高速公路（线段树）

传送门

考虑每一段对答案的贡献

用每一段的左端点来表示当前这一段，那么区间就变成了[1,n-1]

如果询问区间[l,r]，其中一个点的位置为x，则它对答案的贡献为(x-l)*(r-x)*s[x]（s[x]为这一段的权值）

化简后得x*s[x]*(l+r-1)-s[x]*(l*r-r)-x*x*s[x]

那么我们就需要维护x*s[x]，s[x]，x*x*s[x]

其中还需要预处理出来x和x*x

然后就ok了

#include <cstdio>
#include <cstring>
#include <iostream>
#define N 500001
#define LL long long
#define root 1, 1, n - 1
#define ls now << 1, l, mid
#define rs now << 1 | 1, mid + 1, r
 
using namespace std;
 
int n, m;
LL s, xs, xxs, ans1, ans2;
LL x1[N], x2[N], sum1[N], sum2[N], sum3[N], add[N];
//x1表示 x
//x2表示 x^2
//sum1表示 s[x]
//sum2表示 x * s[x]
//sum3表示 x^2 * s[x]
 
inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}
 
inline void push_down(int now, int l, int r)
{
	if(add[now])
	{
		int mid = (l + r) >> 1;
		sum1[now << 1] += 1ll * add[now] * (mid - l + 1);
		sum1[now << 1 | 1] += 1ll * add[now] * (r - mid);
		sum2[now << 1] += add[now] * x1[now << 1];
		sum2[now << 1 | 1] += add[now] * x1[now << 1 | 1];
		sum3[now << 1] += add[now] * x2[now << 1];
		sum3[now << 1 | 1] += add[now] * x2[now << 1 | 1];
		add[now << 1] += add[now];
		add[now << 1 | 1] += add[now];
		add[now] = 0;
	}
}
 
inline void push_up(int now)
{
	sum1[now] = sum1[now << 1] + sum1[now << 1 | 1];
	sum2[now] = sum2[now << 1] + sum2[now << 1 | 1];
	sum3[now] = sum3[now << 1] + sum3[now << 1 | 1];
}
 
inline void update(int now, int l, int r, int x, int y, LL z)
{
	if(x <= l && r <= y)
	{
		add[now] += z;
		sum1[now] += 1ll * z * (r - l + 1);
		sum2[now] += 1ll * z * x1[now];
		sum3[now] += 1ll * z * x2[now];
		return;
	}
	push_down(now, l, r);
	int mid = (l + r) >> 1;
	if(x <= mid) update(ls, x, y, z);
	if(mid < y) update(rs, x, y, z);
	push_up(now);
}
 
inline void build(int now, int l, int r)
{
	if(l == r)
	{
		x1[now] += l;
		x2[now] += 1ll * l * l;
		return;
	}
	int mid = (l + r) >> 1;
	build(ls);
	build(rs);
	x1[now] = x1[now << 1] + x1[now << 1 | 1];
	x2[now] = x2[now << 1] + x2[now << 1 | 1];
}
 
inline void query(int now, int l, int r, int x, int y)
{
	if(x <= l && r <= y)
	{
		s += sum1[now];
		xs += sum2[now];
		xxs += sum3[now];
		return;
	}
	push_down(now, l, r);
	int mid = (l + r) >> 1;
	if(x <= mid) query(ls, x, y);
	if(mid < y) query(rs, x, y);
}
 
inline LL gcd(LL x, LL y)
{
	return !y ? x : gcd(y, x % y);
}
 
int main()
{
	LL g, z;
	int i, x, y;
	char c[10];
	n = read();
	m = read();
	build(root);
	while(m--)
	{
		scanf("%s", c);
		if(c[0] == 'C')
		{
			x = read();
			y = read();
			z = read();
			update(root, x, y - 1, z);
		}
		else
		{
			x = read();
			y = read();
			s = xs = xxs = 0;
			query(root, x, y - 1);
			ans2 = 1ll * (1 + y - x) * (y - x) / 2;
			ans1 = 1ll * xs * (x + y - 1) - s * (1ll * x * y - y) - xxs;
			g = gcd(ans1, ans2);
			printf("%lld/%lld\n", ans1 / g, ans2 / g);
		}
	}
	return 0;
}

　　

一个longlong调了我45min，WNM