CF121E Lucky Array-526互联

题意

给定一个序列，维护下列操作。

区间加
区间查询数中只包含 \(4, 7\) 数的个数。

所有数前后不超过 \(1e4\)。

Sol

块块版。

\(1e4\)，发现满足条件的数的个数只有 \(30\) 个。

对于每个块开一个桶，记录每种数有多少个。

查询时暴力枚举 \(30\) 个数，暴力判断即可。

修改是平凡的。

Code

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <vector>
using namespace std;
#ifdef ONLINE_JUDGE

#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;

#endif
int read() {
	int p = 0, flg = 1;
	char c = getchar();
	while (c < '0' || c > '9') {
		if (c == '-') flg = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9') {
		p = p * 10 + c - '0';
		c = getchar();
	}
	return p * flg;
}
string read_() {
	string ans;
	char c = getchar();
	while (c < 'a' || c > 'z')
		c = getchar();
	while (c >= 'a' && c <= 'z')
		ans += c, c = getchar();
	return ans;
}
void write(int x) {
	if (x < 0) {
		x = -x;
		putchar('-');
	}
	if (x > 9) {
		write(x / 10);
	}
	putchar(x % 10 + '0');
}
const int N = 1e5 + 5, blk = 327;

array <array <int, N>, blk> bk;
array <int, blk> tag;
array <int, N> s;

int calc(int x) {
	return (x - 1) / blk + 1;
}

vector <int> isl;

void modify(int l, int r, int x) {
	if (calc(l) == calc(r)) {
		for (int i = l; i <= r; i++)
			bk[calc(i)][s[i]]--, s[i] += x, bk[calc(i)][s[i]]++;
		return;
	}
	for (int i = l; i <= calc(l) * blk; i++)
		bk[calc(i)][s[i]]--, s[i] += x, bk[calc(i)][s[i]]++;
	for (int i = (calc(r) - 1) * blk + 1; i <= r; i++)
		bk[calc(i)][s[i]]--, s[i] += x, bk[calc(i)][s[i]]++;
	for (int i = calc(l) + 1; i <= calc(r) - 1; i++)
		tag[i] += x;
}

bool check(int x) {
	for (auto y : isl)
		if (x == y) return 1;
	return 0;
}

int query(int l, int r) {
	int ans = 0;
	if (calc(l) == calc(r)) {
		for (int i = l; i <= r; i++)
			if (check(s[i] + tag[calc(l)]))
				ans++;
		return ans;
	}
	for (int i = l; i <= calc(l) * blk; i++)
		ans += check(s[i] + tag[calc(l)]);
	for (int i = (calc(r) - 1) * blk + 1; i <= r; i++)
		ans += check(s[i] + tag[calc(r)]);
	for (int i = calc(l) + 1; i <= calc(r) - 1; i++) {
		for (auto x : isl) {
			if (x - tag[i] <= 0) continue;
			ans += bk[i][x - tag[i]];
		}
	}
	return ans;
}

int main() {
	int n = read(), q = read();
	for (int i = 1; i <= n; i++)
		s[i] = read();
	for (int i = 1; i <= calc(n); i++)
		for (int j = (i - 1) * blk + 1; j <= i * blk; j++)
			bk[i][s[j]]++;
	for (int l = 1; l <= 4; l++) {
		for (int T = 0; T < 1 << l; T++) {
			int idx = 0, tp = 1;
			for (int i = 1; i <= l; i++) {
				idx += (T & (1 << (i - 1)) ? 7 : 4)	* tp;
				tp *= 10;
			}
			isl.push_back(idx);
		}
	}
	/* for (auto x : isl) */
		/* write(x), putchar(32); */
	/* puts(""); */
	while (q--) {
		string tp = read_();
		int l = read(), r = read();
		if (tp == "add") {
			int x = read();
			modify(l, r, x);
		}
		else write(query(l, r)), puts("");
	}

	return 0;
}