三道计数 + 一道数据结构也是没谁了
这场可是好好锻炼了我的写暴搜能力。。。
挤压 20pts
暴搜20pts；
把最后的答案进行二进制拆解，即 $ ans = 2^i + 2^j + 2^k + ... $，那么答案的平方为 $ \sum_{i = 0}^{30} \sum_{j = 0}^{30} s_i s_j 2^{i + j} $，其中 $ s_i, s_j $ 代表答案二进制拆解后这一位是 $ 1 $ 还是 $ 0 $；
那么我们发现，答案的平方只和答案二进制拆解后的任意两位有关，所以我们直接设 $ f_{k, i, j, 0/1, 0/1} $ 表示考虑到第 $ k $ 个数字，第 $ i $ 位是 $ 0/1 $，第 $ j $ 位是 $ 0/1 $ 的概率即可；
这其实是个概率 $ DP $，从第一步开始其实比较难想；
点击查看代码
#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
const long long mod = 1e9 + 7;
long long n;
long long a[500005], q[500005];
long long ksm(long long a, long long b) {long long ans = 1;while(b) {if (b & 1) ans = ans * a % mod;a = a * a % mod;b >>= 1;}return ans;
}
long long f[35][35][2][2], now[2][2];
int main() {freopen("a.in", "r", stdin);freopen("a.out", "w", stdout);ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin >> n;long long kk = ksm(1000000000, mod - 2);for (int i = 1; i <= n; i++) cin >> a[i];for (int i = 1; i <= n; i++) {cin >> q[i];q[i] = q[i] * kk % mod;}for (int i = 0; i <= 30; i++) {for (int j = 0; j <= 30; j++) {f[i][j][0][0] = 1;}}for (int i = 1; i <= n; i++) {for (int j = 0; j <= 30; j++) {for (int k = 0; k <= 30; k++) {for (int l = 0; l <= 1; l++) {for (int r = 0; r <= 1; r++) {now[l][r] = f[j][k][l][r];}}bool tj = false;bool tk = false;if ((a[i] >> j) & 1) tj = true;if ((a[i] >> k) & 1) tk = true;for (int l = 0; l <= 1; l++) {for (int r = 0; r <= 1; r++) {f[j][k][l][r] = (now[l][r] * (1 - q[i] + mod) % mod + now[l ^ tj][r ^ tk] * q[i] % mod) % mod;}}}}}long long ans = 0;for (int i = 0; i <= 30; i++) {for (int j = 0; j <= 30; j++) {ans = (ans + f[i][j][1][1] * ksm(2, i + j) % mod) % mod;}}cout << ans;return 0;
}
工地难题 30pts
暴搜30pts
考虑正解，是个容斥，题解详见[算法] 容斥的例题；
星空遗迹 20pts
$ \Theta(n^3) $ 暴力20pts；
考虑正解，发现可以单调栈维护，满足下面的能够赢上面的，答案就是栈大小最小时对应的元素，于是可以线段树区间修改区间查最小；
时间复杂度：$ \Theta(n \log n) $；
点击查看代码
#include <iostream>
#include <cstdio>
using namespace std;
int n, q;
char s[500005];
char win(char x, char y) {if (x == y) return x;if ((x == 'R' && y == 'S') || (x == 'S' && y == 'R')) return 'R';if ((x == 'S' && y == 'P') || (x == 'P' && y == 'S')) return 'S';if ((x == 'P' && y == 'R') || (x == 'R' && y == 'P')) return 'P';
}
int f[500005];
namespace SEG{inline int ls(int x) {return x << 1;}inline int rs(int x) {return x << 1 | 1;}struct sss{int l, r, lz;pair<int, int> mi;}tr[3000005];inline void push_up(int id) {tr[id].mi = min(tr[ls(id)].mi, tr[rs(id)].mi);}inline void push_down(int id) {if (tr[id].lz) {tr[ls(id)].lz += tr[id].lz;tr[rs(id)].lz += tr[id].lz;tr[ls(id)].mi.first += tr[id].lz;tr[rs(id)].mi.first += tr[id].lz;tr[id].lz = 0;}}void bt(int id, int l, int r) {tr[id].l = l;tr[id].r = r;if (l == r) {tr[id].mi = {f[l], l};return;}int mid = (l + r) >> 1;bt(ls(id), l, mid);bt(rs(id), mid + 1, r);push_up(id);}void add(int id, int l, int r, int d) {if (tr[id].l >= l && tr[id].r <= r) {tr[id].mi.first += d;tr[id].lz += d;return;}push_down(id);int mid = (tr[id].l + tr[id].r) >> 1;if (l <= mid) add(ls(id), l, r, d);if (r > mid) add(rs(id), l, r, d);push_up(id);}pair<int, int> ask(int id, int l, int r) {if (tr[id].l >= l && tr[id].r <= r) {return tr[id].mi;}push_down(id);int mid = (tr[id].l + tr[id].r) >> 1;if (r <= mid) return ask(ls(id), l, r);else if (l > mid) return ask(rs(id), l, r);else return min(ask(ls(id), l, mid), ask(rs(id), mid + 1, r));}
}
using namespace SEG;
int main() {freopen("a.in", "r", stdin);freopen("a.out", "w", stdout);ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin >> n >> q;for (int i = 1; i <= n; i++) {cin >> s[i];}f[1] = 1;for (int i = 2; i <= n; i++) {if (s[i] == s[i - 1]) {f[i] = f[i - 1];continue;}if (win(s[i], s[i - 1]) == s[i]) {f[i] = f[i - 1] - 1;continue;}if (win(s[i], s[i - 1]) == s[i - 1]) {f[i] = f[i - 1] + 1;continue;}}bt(1, 1, n);int ss, l, r;char y;for (int i = 1; i <= q; i++) {cin >> ss >> l;if (ss == 1) {cin >> y;if (y == s[l]) continue;if (l > 1) {int ls = 0;if (s[l - 1] == s[l]) ls = 0;else if (win(s[l - 1], s[l]) == s[l - 1]) ls = 1;else if (win(s[l - 1], s[l]) == s[l]) ls = -1;int now = 0;if (s[l - 1] == y) now = 0;else if (win(s[l - 1], y) == s[l - 1]) now = 1;else if (win(s[l - 1], y) == y) now = -1;add(1, l, n, now - ls);}if (l < n) {int ls = 0;if (s[l + 1] == s[l]) ls = 0;else if (win(s[l], s[l + 1]) == s[l]) ls = 1;else if (win(s[l], s[l + 1]) == s[l + 1]) ls = -1;int now = 0;if (y == s[l + 1]) now = 0;else if (win(s[l + 1], y) == y) now = 1;else if (win(s[l + 1], y) == s[l + 1]) now = -1;add(1, l + 1, n, now - ls); }s[l] = y;}if (ss == 2) {cin >> r;cout << s[ask(1, l, r).second] << '\n';}}return 0;
}
