100 lines
2.4 KiB
C++
100 lines
2.4 KiB
C++
#include <cstdio>
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#include <cstring>
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#include <algorithm>
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#include <iostream>
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using namespace std;
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const int N = 100010;
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#define ls u << 1
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#define rs u << 1 | 1
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struct Node {
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int l, r;
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int add, sum;
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} tr[N << 2];
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void pushup(int u) {
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tr[u].sum = tr[ls].sum | tr[rs].sum; //因为是模拟二进制的操作，所以采用 或 运算即可完成拼接操作
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}
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void pushdown(int u) {
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if (tr[u].add) {
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tr[ls].add = tr[rs].add = tr[u].add;
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tr[ls].sum = tr[rs].sum = tr[u].add;
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tr[u].add = 0;
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}
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}
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void build(int u, int l, int r) {
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tr[u].l = l, tr[u].r = r;
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if (l == r) {
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tr[u].sum = 1; // sum记录的其实是一个30位的二进制压缩模拟数字，此时置为1表示最后一位即0位为1，是默认的颜色
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return;
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}
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int mid = (l + r) >> 1;
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build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
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pushup(u);
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}
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void modify(int u, int l, int r, int c) {
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if (l <= tr[u].l && tr[u].r <= r) {
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tr[u].add = 1 << (c - 1);
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tr[u].sum = 1 << (c - 1);
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return;
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}
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pushdown(u);
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int mid = tr[u].l + tr[u].r >> 1;
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if (l <= mid) modify(u << 1, l, r, c);
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if (r > mid) modify(u << 1 | 1, l, r, c);
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pushup(u);
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}
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int query(int u, int l, int r) {
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if (l <= tr[u].l && tr[u].r <= r) return tr[u].sum;
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int ans = 0;
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pushdown(u);
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int mid = tr[u].l + tr[u].r >> 1;
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if (l <= mid) ans = query(u << 1, l, r);
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if (r > mid) ans = ans | query(u << 1 | 1, l, r);
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return ans;
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}
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//手工版本数字1个数
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int count(int x) {
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int cnt = 0;
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while (x) {
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cnt++;
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x -= (x & -x);
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}
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return cnt;
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}
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int main() {
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//加快读入
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ios::sync_with_stdio(false), cin.tie(0);
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int n, t, m; //区间[1,n],t种颜色,m个操作
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cin >> n >> t >> m;
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//构建
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build(1, 1, n);
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while (m--) {
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char op;
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int l, r;
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cin >> op >> l >> r;
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if (l > r) swap(l, r); //坑点题目没说哪个大哪个小，需要特判
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if (op == 'C') {
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int c;
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cin >> c;
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modify(1, l, r, c);
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} else {
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int ans = query(1, l, r);
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int cnt = count(ans); //获取某个数二进制位1的个数
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printf("%d\n", cnt);
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}
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}
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return 0;
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} |