題目鏈接

BZOJ5322

題解

意思就是使有序的排列盡量少
就是使相同的數盡量少
然後大力亂搞即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
 

#define cp pair<int,int>
#define LL long long int
#define res register
using namespace std;
const int maxn = 200005,maxm = 10200005,INF = 1000000000,P = 998244353;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while
 
 (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int n,m,l,r,a[maxn];
int fac[maxm];
inline int qpow(int a,int b){
    int ans = 1;
    for (; b; b >>= 1,a = 1ll * a * a % P)
        if (b & 1) ans = 1ll * ans * a % P;
    return
 
 ans;
}
void init(){
    fac[0] = 1;
    for (res int i = 1; i < maxm; i++)
        fac[i] = 1ll * fac[i - 1] * i % P;
}
int bac[maxm],tail;
int main(){
    init();
    int T = read(),ans;
    while (T--){
        n = read(); m = read(); l = read(); r = read(); ans = fac[n + m];
        REP(i,n) a[i] = read();
        sort(a + 1,a + 1 + n);
        tail = 0; int cnt = 0,M = 0;
        for (res int i = 1; i <= n; i++){
            if (i != 1 && a[i] != a[i - 1]){
                if (a[i - 1] >= l && a[i - 1] <= r){
                    bac[cnt]++,tail++,M = max(M,cnt);
                }
                else ans = 1ll * ans * qpow(fac[cnt],P - 2) % P;
                cnt = 1;
            }
            else cnt++;
        }
        if (a[n] >= l && a[n] <= r){
            bac[cnt]++,tail++,M = max(M,cnt);
        }
        else ans = 1ll * ans * qpow(fac[cnt],P - 2) % P;
        bac[0] += r - l + 1 - tail;
        for (res int i = 0; m; i++,M = max(M,i)){
            if (M == i){
                int tot = m / bac[i],lef = m - tot * bac[i];
                ans = 1ll * ans * qpow(qpow(fac[i + tot],P - 2),bac[i] - lef) % P;
                ans = 1ll * ans * qpow(qpow(fac[i + tot + 1],P - 2),lef) % P;
                bac[i] = M = 0;
                break;
            }
            if (bac[i] >= m){
                bac[i + 1] += m;
                bac[i] -= m;
                m = 0;
            }
            else {
                m -= bac[i];
                bac[i + 1] += bac[i];
                bac[i] = 0;
            }
        }
        for (res int i = 0; i <= M; i++){
            ans = 1ll * ans * qpow(qpow(fac[i],P - 2),bac[i]) % P;
            bac[i] = 0;
        }
        printf("%d\n",ans);
    }
    return 0;
}

BZOJ5322 [Jxoi2018]排序問題 【亂搞】