Codeforces Round #684 (Div. 1)题解

A1/A2.Binary Table

码农题，其实封装一下函数就不用写很长了

每次操作可以使一个1变成0

我们可以先用$m(n-2)$次操作将$([1,1],[n-2,m])$变为$0$

然后再用$m-1$次操作将$([n-1,1],[n,m-1])$变为$0$

于是只剩下$(n-1,m)$和$(n,m)$，三下操作可以都变为$0$

最多$(n-1)m+2$次，实际操作中会出现很多

代码太丑，就不放了

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C.Greedy Shopping

Part 1：

线段树二分

操作1就是二分找到一个最大的小于$y$的值，区间覆盖，时间复杂度$O(\log n)$

操作2就每次二分一个起点，再二分一个终点，容易发现操作区间数量不超过$\log n$，时间复杂度为$O(\log^2 n)$

但是起点限制较为麻烦，参考$\tt{K\color{red}{arry5307}}$的题解，先将$y$加上$\begin{matrix} \sum_{i=1}^{x-1} a[i] \end{matrix}$，然后就可以以$1$为起点了

总时间复杂度为$O(n \log^2 n)$

Part 2:

对于2操作中区间数量不超过$\log n$的证明：

因为序列单调递减，所以如果$y=y-a[i]$后仍大于原来的$\dfrac{y}{2}$，它可以继续减下去

于是减了一个区间$y$至少缩小一半，区间数量$\le$$\log n$

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int n, q, opt, x;
ll a[200011], y;
struct segment_tree {
	ll minn[800011], sum[800011], tag[800011];
	void build(int k, int l, int r) {
		if(l == r) {
			sum[k] = minn[k] = a[l];
			return;
		}
		int mid = (l + r) >> 1;
		build(k<<1, l, mid);
		build(k<<1|1, mid+1, r);
		minn[k] = min(minn[k<<1], minn[k<<1|1]);
		sum[k] = sum[k<<1] + sum[k<<1|1];
	}
	void pushdown(int k, int l, int r) {
		if(tag[k]) {
			int mid = (l + r) >> 1;
			tag[k<<1] = tag[k<<1|1] = tag[k];
			minn[k<<1] = minn[k<<1|1] = tag[k];
			sum[k<<1] = (mid-l+1)*tag[k];
			sum[k<<1|1] = (r-mid)*tag[k];
			tag[k] = 0;
		}
	}
	void modify(int k, int l, int r, int L, int R, ll num) {
		if(l >= L && r <= R) {
			minn[k] = num;
			tag[k] = num;
			sum[k] = (r - l + 1) * num;
			return;
		}
		pushdown(k, l, r);
		int mid = (l + r) >> 1;
		if(L <= mid) modify(k<<1, l, mid, L, R, num);
		if(R > mid) modify(k<<1|1, mid+1, r, L, R, num);
		minn[k] = min(minn[k<<1], minn[k<<1|1]);
		sum[k] = sum[k<<1] + sum[k<<1|1];
	} //区间覆盖 
	int binary(int k, int l, int r, ll num) {
		if(l == r) {
			if(minn[k] <= num) return l;
			return n+1;
		} 
		pushdown(k, l, r);
		int mid = (l + r) >> 1;
		if(minn[k<<1] <= num) return binary(k<<1, l, mid, num);
		if(minn[k<<1|1] <= num) return binary(k<<1|1, mid+1, r, num);
		return n+1;
	} //二分查找最左边的<=num的下标 
	int binary1(int k, int l, int r, ll num) {
		if(sum[k] <= num) return n+1;
		if(l == r) return l;
		pushdown(k, l, r);
		int mid = (l + r) >> 1;
		if(sum[k<<1] > num) return binary1(k<<1, l, mid, num);
		return binary1(k<<1|1, mid+1, r, num-sum[k<<1]); 
	} //二分查找最左边的前缀和>num的坐标 
	ll query(int k, int l, int r, int L, int R) {
		if(L > R) return 0;
		if(l >= L && r <= R) return sum[k];
		pushdown(k, l, r);
		int mid = (l + r) >> 1; ll ret = 0;
		if(L <= mid) ret += query(k<<1, l, mid, L, R);
		if(R > mid) ret += query(k<<1|1, mid+1, r, L, R);
		return ret; 
	} //区间求和 
} t;
int main() {
	scanf("%d%d", &n, &q);
	for(int i = 1; i <= n; i++) scanf("%lld", &a[i]);
	t.build(1, 1, n);
	for(int i = 1; i <= q; i++) {
		scanf("%d%d%lld", &opt, &x, &y);
		if(opt == 1) {
			int w = t.binary(1, 1, n, y);
			if(w > x) continue;
			t.modify(1, 1, n, w, x, y);
		}
		else {
			y += t.query(1, 1, n, 1, x-1);
			int w = t.binary(1, 1, n, y), ans = -(x-1);
			while(w != n+1) {
				ll sum = t.query(1, 1, n, 1, w-1);
				int r = t.binary1(1, 1, n, y+sum) - 1;
				ans += (r - w + 1);
				if(r == n) break;
				y -= t.query(1, 1, n, w, r);
				w = t.binary(1, 1, n, y);
			} 
			printf("%d\n", ans);
		}
	}
	return 0;
}