#6029. 「雅礼集训 2017 Day1」市场 （势能线段树）

传送门

PS:这题正数与负数都需要向下取整，也就是说-4 / 3 = -2

解题思路：

  很玄学的写法

  线段树上维护最大值max和最小值min，可以发现对于除法而言(max / x) - (min / x)会导致max和min的差值越来越小，这个可以理解为势能的减小。对于一个区间而言，若max和min符号相同 且abs(max - max / x) == abs(min - min / x)那么区间所有数字的除法可以看做减法，都减去abs(max - max/ x)，但是还有add操作，这个可能会导致势能增加，不太懂为什么这样就能过，感觉很玄学。

---

  

#include <bits/stdc++.h>

using ll = long long;
const int N = 1e5 + 10;
const int MOD = 1e9 + 7;
const ll INF = 0x3f3f3f3f3f3f3f3f * 2;


typedef std::pair<int, int> PII;
int n, m;

int a[N];

ll mn[N << 2], mx[N << 2];
ll sum[N << 2], add[N << 2];

#define ls u << 1
#define rs u << 1 | 1

inline void pushup(int u) {
	sum[u] = sum[u << 1] + sum[u << 1 | 1];
	mn[u] = std::min(mn[ls], mn[rs]);
	mx[u] = std::max(mx[ls], mx[rs]);
}

inline void pushdown(int u, int L, int R) {
	if (add[u]) {
		int mid = L + R >> 1;
		add[ls] += add[u];
		add[rs] += add[u];
		sum[ls] += 1ll * (mid - L + 1) * add[u];
		sum[rs] += 1ll * (R - mid) * add[u];
		mn[ls] += add[u];
		mn[rs] += add[u];
		mx[ls] += add[u];
		mx[rs] += add[u];
		add[u] = 0;
	}
}

inline void build(int u, int l, int r) {
	if (l == r) {
		sum[u] = mx[u] = mn[u] = a[l];
		return ;
	}
	
	int mid = l + r >> 1;
	build(ls, l, mid);
	build(rs, mid + 1, r);
	
	pushup(u);
}

inline void DIV(int u, int L, int R, int l, int r, int x) {
	if (L >= l && R <= r) {
		if (mx[u] >= 0 && mn[u] >= 0) {
			ll t1 = mx[u] / x, t2 = mn[u] / x;
			if (abs(mx[u] - t1) == abs(mn[u] - t2)) {
				sum[u] -= 1ll * (R - L + 1) * abs(mx[u] - t1);
				add[u] -= abs(t1 - mx[u]);
				mn[u] -= abs(mx[u] - t1);
				mx[u] -= abs(mx[u] - t1);
			} else {
				pushdown(u, L, R);
				int mid = L + R >> 1;
	
				if (l <= mid) DIV(ls, L, mid, l, r, x);
				if (r > mid) DIV(rs, mid + 1, R, l, r, x);
				
				pushup(u);
			}
		} else if (mx[u] <= 0 && mn[u] <= 0){
			ll t1 = (mx[u] - x + 1) / x, t2 = (mn[u] - x + 1) / x;
			if (abs(mx[u] - t1) == abs(mn[u] - t2)) {
				sum[u] += 1ll * (R - L + 1) * abs(mx[u] - t1);
				add[u] += abs(t1 - mx[u]);
		
				mn[u] += abs(mx[u] - t1);
				mx[u] += abs(mx[u] - t1);
				
			} else {
				pushdown(u, L, R);
				int mid = L + R >> 1;
	
				if (l <= mid) DIV(ls, L, mid, l, r, x);
				if (r > mid) DIV(rs, mid + 1, R, l, r, x);
				
				pushup(u);
			}
		} else {
			pushdown(u, L, R);
			int mid = L + R >> 1;

			if (l <= mid) DIV(ls, L, mid, l, r, x);
			if (r > mid) DIV(rs, mid + 1, R, l, r, x);
			
			pushup(u);
		}
		
		return ;
	}
	
	pushdown(u, L, R);
	
	int mid = L + R >> 1;
	
	if (l <= mid) DIV(ls, L, mid, l, r, x);
	if (r > mid) DIV(rs, mid + 1, R, l, r, x);
	
	pushup(u);
}

inline void ADD(int u, int L, int R, int l, int r, int x) {
	if (L >= l && R <= r) {
		add[u] += x;
		sum[u] += 1ll * (R - L + 1) * x;
		mn[u] += x;
		mx[u] += x;
		return ;
	}
	
	pushdown(u, L, R);
	int mid = L + R >> 1;
	
	if (l <= mid) ADD(ls, L, mid, l, r, x);
	if (r > mid) ADD(rs, mid + 1, R, l, r, x);
	
	pushup(u);
}

inline ll querymin(int u, int L, int R, int l, int r) {
	if (L >= l && R <= r) return mn[u];
	
	pushdown(u, L, R);
	int mid = L + R >> 1;
	ll ans = INF;
	if (l <= mid) ans = querymin(ls, L, mid, l, r);
	if (r > mid) ans = std::min(ans, querymin(rs, mid + 1, R, l, r));
	
	pushup(u);
	return ans;
}

inline ll querysum(int u, int L, int R, int l, int r) {
	if (L >= l && R <= r) return sum[u];
	
	pushdown(u, L, R);
	int mid = L + R >> 1;
	ll ans = 0;
	if (l <= mid) ans = querysum(ls, L, mid, l, r);
	if (r > mid) ans += querysum(rs, mid + 1, R, l, r);

	pushup(u);
	return ans;
}

inline void solve() {
	std::cin >> n >> m;
	
	for (int i = 1; i <= n; i ++) 
		std::cin >> a[i];
		
	build(1, 1, n);
		
	while (m --) {
		int op, l, r, x;
		std::cin >> op >> l >> r;
		l ++, r ++;
		if (op == 1) {
			std::cin >> x;
			ADD(1, 1, n, l, r, x);
		} else if (op == 2) {
			std::cin >> x;
			DIV(1, 1, n, l, r, x);
		} else if (op == 3) {
			std::cout << querymin(1, 1, n, l, r) << '\n';
		} else {
			std::cout << querysum(1, 1, n, l, r) << '\n';
		}
	}
}


int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);

    int _ = 1;
    //std::cin >> _;
    while (_ --) solve();
    
    return 0;
}