BZOJ5322 [Jxoi2018]排序问题 【贪心】

题目链接

BZOJ5322

题解

意思就是使有序的排列尽量少
就是使相同的数尽量少
然后大力贪心即可

#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
",ans);
	}
	return 0;
}