洛谷P3600随机数生成器——期望+DP

原题链接
写到一半发现写不下去了。。。
所以orz xyz32768，您去看这篇题解吧，思路很清晰，我之前写的胡言乱语与之差距不啻天渊

#include <algorithm>
#include  <iostream>
#include   <cstdlib>
#include   <cstring>
#include    <cstdio>
#include    <random>
#include    <string>
#include    <vector>
#include     <cmath>
#include     <ctime>
#include     <queue>
#include       <map>
#include       <set>

#define IINF 0x3f3f3f3f3f3f3f3fLL
#define u64 unsigned long long
#define pii pair<int, int>
#define mii map<int, int>
#define u32 unsigned int
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back
#define is insert
#define se second
#define fi first
#define ps push

using namespace std;

#define MOD 666623333

void add(int &x, int y) {
    x = (x + y) % MOD;
    if (x < 0) x += MOD;
}

int Add(int x, int y) {
    return (x + y + MOD) % MOD;
}

void mul(int &x, int y) {
    x = 1LL * x * y % MOD;
    if (x < 0) x += MOD;
}

int Mul(int x, int y) {
    return (1LL * x * y % MOD + MOD) % MOD;
}

int fpow(int x, int p) {
    int ret = 1;
    while(p) {
        if (p & 1) mul(ret, x);
        mul(x, x);
        p >>= 1;
    }
    return ret;
}

const int MAXN = 2000;
int n, x, q, h[MAXN + 5], g[MAXN + 5], f[MAXN + 5][MAXN + 5], sum[MAXN + 5][MAXN + 5], l[MAXN + 5], r[MAXN + 5], tmp[MAXN + 5], ans;
pii qrs0[MAXN + 5], qrs[MAXN + 5];

void init() {
    cin >> n >> x >> q;
    for (int i = 1; i <= q; ++i)
        cin >> qrs0[i].fi >> qrs0[i].se;
    int tot = 0;
    sort(qrs0 + 1, qrs0 + q + 1);
    for (int i = 1; i <= q; ++i) {
        if (i > 1 && qrs0[i].fi == qrs0[i - 1].fi) continue;
        while (tot && qrs[tot].se >= qrs0[i].se) tot--;
        qrs[++tot] = qrs0[i];
    }
    q = tot;
    for (int i = 1; i <= q; ++i) tmp[qrs[i].se + 1]--, tmp[qrs[i].fi]++;
    for (int i = 2; i <= n; ++i) tmp[i] += tmp[i - 1];
    for (int i = 1; i <= n; ++i) {
        if (!tmp[i]) {
            int j = 0;
            while (j + 1 <= q && qrs[j + 1].se < i) j++;
            r[i] = j, l[i] = j + 1;
        }
        else {
            int j = 0;
            while (j + 1 <= q && qrs[j + 1].se < i) j++;
            l[i] = ++j;
            while (j + 1 <= q && qrs[j + 1].fi <= i) j++;
            r[i] = j;
        }
    }
}

void solve() {
    f[0][0] = sum[0][0] = 1;
    int p = -1;
    for (int i = 1; i <= n; ++i) {
        while (p < i && r[p + 1] + 1 < l[i]) p++;
        sum[0][i] = 1;
        for (int j = 1; j <= n; ++j) {
            f[i][j] = Add(sum[j - 1][i - 1], (~p ? -sum[j - 1][p] : 0));
            sum[j][i] = Add(sum[j][i - 1], f[i][j]);
        }
        if (r[i] == q)
            for (int j = 1; j <= n; ++j)
                add(g[j], f[i][j]);
    }
    for (int i = 1; i <= x; ++i)
        for (int j = 1; j <= n; ++j)
            add (h[i], Mul(g[j], Mul(fpow(i, j), fpow(x - i, n - j))));
    for (int i = 1; i <= x; ++i)
        add (ans, Mul(i, Add(h[i], -h[i - 1])));
    mul(ans, fpow(fpow(x, n), MOD - 2));
}

int main() {
    init();
    solve();
    cout << ans << endl;
    return 0;
}