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GMOJ S5020 【牛奶装瓶】

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Description

给你 $n$ 个数，$m$ 个操作，支持区间加、区间除然后向下取整、区间最小值、区间和。

Solution

除了 「区间除然后向下取整」 这个操作比较难处理以外，其他操作都可以简单地用线段树来维护。

考虑如果当前递归到一个区间，区间内最大值和最小值减少的值一样，就可以把原操作变为区间减了。

这个的时间复杂度可证明为 $O(n \log n \log C)$，其中 $C = 10^9\text{（值域）} + 10^4 q$，但我并不会证明。

Code

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#include <cmath>
#include <cstdio>
#define maxN 100010
struct tree{ long long min, max, sum, tag; } a[maxN << 2];
const long long inf = (1LL << 60LL);
long long A[maxN];
long long Min (long long x, long long y)
{
	return x < y ? x : y;
}
long long Max (long long x, long long y)
{
	return x > y ? x : y;
}
void build (long long now, long long l, long long r)
{
	a[now].min = inf;
	a[now].max = -inf;
	a[now].sum = a[now].tag = 0;
	if(l != r)
	{
		long long mid = (l + r) >> 1;
		build(now << 1, l, mid);
		build(now << 1 | 1, mid + 1, r);
		a[now].sum = a[now << 1].sum + a[now << 1 | 1].sum;
		a[now].min = Min(a[now << 1].min, a[now << 1 | 1].min);
		a[now].max = Max(a[now << 1].max, a[now << 1 | 1].max);
	}
	else
	{
		a[now].min = a[now].max = a[now].sum = A[l];
	}
}
void pushdown (long long now, long long l, long long r)
{
	if(!a[now].tag || l == r)
	{
		return ;
	}
	long long mid = (l + r) >> 1;
	a[now << 1].min += a[now].tag;
	a[now << 1].max += a[now].tag;
	a[now << 1].sum += (mid - l + 1) * a[now].tag;
	a[now << 1].tag += a[now].tag;
	a[now << 1 | 1].min += a[now].tag;
	a[now << 1 | 1].max += a[now].tag;
	a[now << 1 | 1].sum += (r - mid) * a[now].tag;
	a[now << 1 | 1].tag += a[now].tag;
	a[now].tag = 0;
}
void change (long long now, long long l, long long r, long long L, long long R, long long x)
{
	if(l == L && r == R)
	{
		a[now].min += x;
		a[now].max += x;
		a[now].sum += (r - l + 1) * x;
		a[now].tag += x;
		return ;
	}
	pushdown(now, l, r);
	long long mid = (l + r) >> 1;
	if(R <= mid)
	{
		change(now << 1, l, mid, L, R, x);
	}
	else if(mid + 1 <= L)
	{
		change(now << 1 | 1, mid + 1, r, L, R, x);
	}
	else
	{
		change(now << 1, l, mid, L, mid, x);
		change(now << 1 | 1, mid + 1, r, mid + 1, R, x);
	}
	a[now].sum = a[now << 1].sum + a[now << 1 | 1].sum;
	a[now].min = Min(a[now << 1].min, a[now << 1 | 1].min);
	a[now].max = Max(a[now << 1].max, a[now << 1 | 1].max);
}
void work (long long now, long long l, long long r, long long L, long long R, long long d)
{
	long long A = (long long)floor(a[now].min * 1.0 / d) - a[now].min;
	long long B = (long long)floor(a[now].max * 1.0 / d) - a[now].max;
	if(A == B && l == L && r == R)
	{
		a[now].min += A;
		a[now].max += A;
		a[now].sum += (r - l + 1) * A;
		a[now].tag += A;
		return ;
	}
	pushdown(now, l, r);
	long long mid = (l + r) >> 1;
	if(R <= mid)
	{
		work(now << 1, l, mid, L, R, d);
	}
	else if(mid + 1 <= L)
	{
		work(now << 1 | 1, mid + 1, r, L, R, d);
	}
	else
	{
		work(now << 1, l, mid, L, mid, d);
		work(now << 1 | 1, mid + 1, r, mid + 1, R, d);
	}
	a[now].sum = a[now << 1].sum + a[now << 1 | 1].sum;
	a[now].min = Min(a[now << 1].min, a[now << 1 | 1].min);
	a[now].max = Max(a[now << 1].max, a[now << 1 | 1].max);
}
long long querymin (long long now, long long l, long long r, long long L, long long R)
{
	if(l == L && r == R)
	{
		return a[now].min;
	}
	pushdown(now, l, r);
	long long mid = (l + r) >> 1;
	if(R <= mid)
	{
		return querymin(now << 1, l, mid, L, R);
	}
	else if(mid + 1 <= L)
	{
		return querymin(now << 1 | 1, mid + 1, r, L, R);
	}
	else
	{
		long long A = querymin(now << 1, l, mid, L, mid);
		long long B = querymin(now << 1 | 1, mid + 1, r, mid + 1, R);
		return Min(A, B);
	}
}
long long querysum (long long now, long long l, long long r, long long L, long long R)
{
	if(l == L && r == R)
	{
		return a[now].sum;
	}
	pushdown(now, l, r);
	long long mid = (l + r) >> 1;
	if(R <= mid)
	{
		return querysum(now << 1, l, mid, L, R);
	}
	else if(mid + 1 <= L)
	{
		return querysum(now << 1 | 1, mid + 1, r, L, R);
	}
	else
	{
		long long A = querysum(now << 1, l, mid, L, mid);
		long long B = querysum(now << 1 | 1, mid + 1, r, mid + 1, R);
		return A + B;
	}
}
int main ()
{
	freopen("milk.in", "r", stdin);
	freopen("milk.out", "w", stdout);
	long long n = 0, m = 0;
	scanf("%lld %lld", &n, &m);
	for(long long i = 1;i <= n; i++)
	{
		scanf("%lld", &A[i]);
	}
	build(1, 1, n);
	while(m--)
	{
		long long t = 0, l = 0, r = 0, x = 0;
		scanf("%lld %lld %lld", &t, &l, &r);
		if(t == 1)
		{
			scanf("%lld", &x);
			change(1, 1, n, l, r, x);
		}
		else if(t == 2)
		{
			scanf("%lld", &x);
			work(1, 1, n, l, r, x);
		}
		else if(t == 3)
		{
			printf("%lld\n", querymin(1, 1, n, l, r));
		}
		else
		{
			printf("%lld\n", querysum(1, 1, n, l, r));
		}
	}
	return 0;
}