洛谷P4561 [JXOI2018]排序问题(二分 期望)

#include<bits/stdc++.h>
#define Fin(x) freopen(#x".in", "r", stdin);
#define int long long 
using namespace std;
const int MAXN = 2e7 + 10, mod = 998244353;
template<typename A, typename B> inline bool chmax(A &x, B y) {return x < y ? x = y, 1 : 0;}
template<typename A, typename B> inline bool chmin(A &x, B y) {return x > y ? x = y, 1 : 0;}
template<typename A, typename B> inline A mul(A x, B y) {return 1ll * x * y % mod;}
template<typename A, typename B> inline void add2(A &x, B y) {x = x + y >= mod ? x + y - mod : x + y;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, fac[MAXN], L, R, a[MAXN], date[MAXN], num, cnt[MAXN], pl, pr, l, r;
int fp(int a, int p) {
    int base = 1;
    while(p) {
        if(p & 1) base = mul(base, a);
        a = mul(a, a);  p >>= 1;
    }
    return base;
}
int inv(int x) {
    return fp(x, mod - 2);
}
int check(int lim) {
    int tot = 0;
    tot = lim * (R - L - (pr - pl));
    if(tot > M) return tot;
    for(int i = pl; i <= pr && tot <= M; i++)
        tot += max(lim - cnt[i], 0ll);
    return tot;
}
void solve() {
    N = read(); M = read(); L = read(); R = read();
    num = 0;
    for(int i = 1; i <= N; i++) a[i] = read(), date[++num] = a[i], cnt[i] = 0;
    sort(a + 1, a + N + 1);
    sort(date + 1, date + num + 1);
    num = unique(date + 1, date + num + 1) - date - 1;
    for(int i = 1; i <= N; i++) a[i] = lower_bound(date + 1, date + num + 1, a[i]) - date, cnt[a[i]]++;
    pl = lower_bound(date + 1, date + num + 1, L) - date;
    pr = upper_bound(date + 1, date + num + 1, R) - date - 1;
    l = 0, r = N + M;
    while(l < r) {
        int mid = l + r >> 1;
        if(check(mid) <= M) l = mid + 1;
        else r = mid;
    }
    int ans = fp(fac[l - 1], R - L - (pr - pl));
    for(int i = 1; i <= num; i++)
        if(i >= pl && i <= pr) ans = mul(ans, fac[max(cnt[i], l - 1)]);
        else ans = mul(ans, fac[cnt[i]]);
    ans = mul(ans, fp(l, M - check(l - 1)));//把少了的补上
    cout << mul(fac[N + M], inv(ans)) <<  '\n';
}
signed main() {
    fac[0] = 1;
    for(int i = 1; i <= (int) 2e7; i++) fac[i] = mul(i, fac[i - 1]);
    for(int T = read(); T--;  solve());
    return 0;
}
/*
2
3 3 5 7
1 3 4
3 3 1 2
1 3 4
*/