[洛谷P2801]教主的魔法

题目大意：有$n$个数，$q$个操作。两种操作：

1. $M\;l\;r\;w：$把$[l,r]$所有数加上$w$
2. $A\;l\;r\;c：$查询$[l,r]$内大于等于$c$的元素的个数。

题解：分块，对于在整块修改改$tag$，非整块暴力修改，查询整块用$lower\_bound$，非整块暴力

卡点：无

C++ Code：

#include <cstdio>
#include <cmath>
#include <algorithm>
#define maxn 1000010
int n, Q, bsz, num;
int l[1010], r[1010], tg[1010];
struct node {
	int w, id;
	inline bool operator < (const node &rhs) const {return w < rhs.w;}
} s[maxn];
int main() {
	scanf("%d%d", &n, &Q); bsz = sqrt(n);
	for (int i = 1; i <= n; i++) scanf("%d", &s[i].w), s[i].id = i;
	num = n / bsz + 1;
	for (int i = 1; i <= num; i++) {
		l[i] = (i - 1) * bsz;
		if (i == 1) l[i] = 1;
		r[i] = i * bsz - 1;
		if (i == num) r[i] = n;
		std::sort(s + l[i], s + r[i] + 1);
	}
	while (Q --> 0) {
		char op[10];
		int x, y, z;
		scanf("%s%d%d%d", op, &x, &y, &z);
		int lb = x / bsz + 1, rb = y / bsz + 1;
		if (op[0] == 'M') {
			if (lb == rb) {
				for (int i = l[lb]; i <= r[lb]; i++) {
					s[i].w = tg[lb];
					if (s[i].id >= x && s[i].id <= y) s[i].w += z;
				}
				tg[lb] = 0;
				std::sort(s + l[lb], s + r[lb]);
			} else {
				for (int i = l[lb]; i <= r[lb]; i++) {
					s[i].w += tg[lb];
					if (s[i].id >= x) s[i].w += z;
				}
				std::sort(s + l[lb], s + r[lb] + 1);
				for (int i = lb + 1; i < rb; i++) tg[i] += z;
				for (int i = l[rb]; i <= r[rb]; i++) {
					s[i].w += tg[lb];
					if (s[i].id <= y) s[i].w += z;
				}
				std::sort(s + l[rb], s + r[rb] + 1);
				tg[lb] = tg[rb] = 0;
			}
		} else {
			int ans = 0;
			if (lb == rb) {
				for (int i = l[lb]; i <= r[lb]; i++) {
					s[i].w += tg[lb];
					if (s[i].id >= x && s[i].id <= y && s[i].w >= z) ans++;
				}
				tg[lb] = 0;
			} else {
				for (int i = l[lb]; i <= r[lb]; i++) {
					s[i].w += tg[lb];
					if (s[i].id >= x && s[i].w >= z) ans++;
				}
				for (int i = lb + 1; i < rb; i++) {
					ans += s + r[i] - std::lower_bound(s + l[i], s + r[i] + 1, (node) {z - tg[i], 0}) + 1;
				}
				for (int i = l[rb]; i <= r[rb]; i++) {
					s[i].w += tg[rb];
					if (s[i].id <= y && s[i].w >= z) ans++;
				}
				tg[lb] = tg[rb] = 0;
			}
			printf("%d\n", ans);
		}
	}
	return 0;
}