BZOJ 2752: [HAOI2012]高速公路(road)

看到期望就想期望的线性性
即求一条边在一个区间里经过多少次
把边权都转到左边的点上
查询和修改就都变成了对 \([l, r - 1]\) 的查询和修改
令 \(x=l,y=r-1\)，那么一条边 \(i\) 在区间 \([x,y]\) 里的贡献为 $$(i - x + 1)(y- i + 1)=(x+y)i \times v_i- i^2 \times v_i+(y-x-xy+1)\times v_x$$
就维护 \(v_i\)，\(i\times v_i\)，\(i^2 \times v_i\) 即可

#include <bits/stdc++.h>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) >> 1)
#define ll long long

const int N = 1e5 + 7;
ll sum[N][3];

struct Seg {
	ll s[N << 2][3];
	int lazy[N << 2];
	inline void tag(int p, int v, int l, int r) {
		lazy[p] += v;
		for (int i = 0; i < 3; i++)
			s[p][i] += (sum[r][i] - sum[l - 1][i]) * v;
	}
	inline void pushdown(int p, int l, int r) {
		if (!lazy[p]) return;
		tag(lp, lazy[p], l, mid);
		tag(rp, lazy[p], mid + 1, r);
		lazy[p] = 0;
	}
	inline void pushup(int p) {
		for (int i = 0; i < 3; i++)
			s[p][i] = s[lp][i] + s[rp][i];
	}
	void update(int p, int l, int r, int x, int y, int v) {
		if (x <= l && y >= r) return tag(p, v, l, r);
		pushdown(p, l, r);
		if (x <= mid) update(lp, l, mid, x, y, v);
		if (y > mid) update(rp, mid + 1, r, x, y, v);
		pushup(p);
	}
	ll query(int p, int l, int r, int x, int y) {
		if (x <= l && y >= r) return 1LL * (x + y) * s[p][1] - 1LL * s[p][2] + (y - x - 1LL * y * x + 1LL) * s[p][0]; 
		pushdown(p, l, r);
		ll ans = 0;
		if (x <= mid) ans += query(lp, l, mid, x, y);
		if (y > mid) ans += query(rp, mid + 1, r, x, y);
		return ans;
	}
} seg;

inline ll C(int n) {
	return 1LL * n * (n - 1) / 2;
}

ll gcd(ll a, ll b) {
	while (b) {
		ll t = a % b;
		a = b;
		b = t;
	}
	return a;
}

int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		sum[i][0] = sum[i - 1][0] + 1;
		sum[i][1] = sum[i - 1][1] + i;
		sum[i][2] = sum[i - 1][2] + 1LL * i * i;
	}
	n--;
	for (int l, r, v; m--; ) {
		char s[3];
		scanf("%s%d%d", s, &l, &r);
		r--;
		if (s[0] == 'C') {
			scanf("%d", &v);
			seg.update(1, 1, n, l, r, v);
		} else {
			ll fz = seg.query(1, 1, n, l, r);
			ll fm = C(r - l + 2);
			ll g = gcd(fz, fm);
			printf("%lld/%lld\n", fz / g, fm / g);
		}
	}
	return 0;
}