牛客练习赛69 D-火柴排队 dp + 组合
传送门: https://ac.nowcoder.com/acm/contest/7329/D
题意
给长度为n的序列和一个d,分别在n个数中选k(1<=k<=n)个数,将这k个数增加d之后不影响排名,要求 a i + d < = a j ai+d<=aj ai+d<=aj。
分别输出每一个k的概率。
思路
要是想在每个k中找出合法的方案,不管是正向的DFS还是反向的容斥,这都会导致T掉(主要是不会做,做了也会错)。看到n=5000时,O(n^2)的复杂度是可以的,那就可以考虑dp了。
设dp[i][j][1/0]表示前i个人中选了j个增加d,并且当前这一位选1/不选0。
所以就有两种选择,选或不选。
当前这一位没有被选,状态转移就是:
d p [ i ] [ j ] [ 0 ] = d p [ i − 1 ] [ j − 1 ] [ 0 ] + d p [ i − 1 ] [ j − 1 ] [ 1 ] ∗ ( a [ i − 1 ] + d < = a [ i ] ) dp[i][j][0] = dp[i - 1][j - 1][0] + dp[i - 1][j - 1][1] * (a[i - 1] + d <= a[i] ) dp[i][j][0]=dp[i−1][j−1][0]+dp[i−1][j−1][1]∗(a[i−1]+d<=a[i])
如果上一位被选了,这一就要判断上一位选了之后增加的d是否合法,合法就可要,不合法就不要。
当前这一位被选,状态转移就是:
d p [ i ] [ j ] [ 1 ] = d p [ i − 1 ] [ j − 1 ] [ 0 ] + d p [ i − 1 ] [ j − 1 ] [ 1 ] dp[i][j][1]=dp[i - 1][j- 1][0] + dp[i - 1][j-1][1] dp[i][j][1]=dp[i−1][j−1][0]+dp[i−1][j−1][1]
由于n=5000,dp[5000][5000][2]可能会爆掉,那么就需要用到滚动数组(因为转移只会与上一行有关)。
设 n o w = i % 2 , p r e = n o w ∧ 1 , 则 状 态 转 移 为 : 设now=i\%2,pre =now \wedge 1,则状态转移为: 设now=i%2,pre=now∧1,则状态转移为:
f [ n o w ] [ j ] [ 0 ] = f [ p r e ] [ j ] [ 0 ] + f [ p r e ] [ j ] [ 1 ] ∗ ( a [ i − 1 ] + d < = a [ i ] ) f[now][j][0] = f[pre][j][0] + f[pre][j][1] * (a[i - 1] + d <= a[i]) f[now][j][0]=f[pre][j][0]+f[pre][j][1]∗(a[i−1]+d<=a[i])
f [ n o w ] [ j ] [ 1 ] = f [ p r e ] [ j − 1 ] [ 0 ] + f [ p r e ] [ j − 1 ] [ 1 ] f[now][j][1] = f[pre][j - 1][0] + f[pre][j - 1][1] f[now][j][1]=f[pre][j−1][0]+f[pre][j−1][1]
最后,预处理所有的阶乘、逆元、阶乘逆元。
Code(64ms)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pdd;
#define INF 0x3f3f3f
#define lc t[u].ls
#define rc t[u].rs
#define m (l + r) / 2
#define lowbit(x) (x & -x)
#define mid (t[u].l + t[u].r) / 2
#define mem(a,b) memset(a , b , sizeof(a))
#define FOR(i, x, n) for(int i = x;i <= n; i++)
// const ll mod = 1e9 + 7;
const ll mod = 998244353;
// const double pi = acos(-1);
// const double eps = 1e-6;
const int N = 5005;
ll quick_pow(ll a, ll b) {
ll ans = 1;
while(b) {
if(b & 1) ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans % mod;
}
ll F[N + 10]; // 阶乘
ll invF[N + 10]; // 阶乘逆元
ll invn[N + 10];
ll a[N + 10];
int vis[N + 10];
ll f[2][N][2];
inline void Init()
{
F[0] = F[1] = invF[0] = invF[1] = invn[0] = invn[1] = 1;
for (int i = 2; i < N; i++)
{
F[i] = F[i - 1] * i % mod;
invn[i] = (mod - mod / i) * invn[mod % i] % mod;
invF[i] = invF[i - 1] * invn[i] % mod;
}
}
inline ll getC(int n, int k)
{
if (k < 0 || n < 0 || n > k)
return 0;
ll ans = F[k];
ans = ans * invF[n] % mod;
ans = ans * invF[k - n] % mod;
return ans;
}
void solve()
{
Init();
int n, d;
cin >> n >> d;
for(int i = 1;i <= n; i++) cin >> a[i];
sort(a + 1, a + n + 1);
a[n + 1] = a[n] + d * 2;
f[1][1][1] = 1;
f[1][0][0] = 1;
for(int i = 2;i <= n; i++) {
int now = i % 2, pre = now ^ 1;
mem(f[now], 0);
for(int j = 0;j <= i; j++) {
f[now][j][0] = (f[pre][j][0] + f[pre][j][1] * (a[i - 1] + d <= a[i])) % mod;
f[now][j][1] = (f[pre][j - 1][0] + f[pre][j - 1][1]) % mod;
}
}
int now = n % 2;
for(int i = 1;i <= n; i++) {
cout << (f[now][i][0] + f[now][i][1]) * quick_pow(getC(i, n), mod - 2) % mod << endl;
}
}
signed main() {
ios_base::sync_with_stdio(false);
// cin.tie(nullptr);
// cout.tie(nullptr);
#ifdef FZT_ACM_LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
signed test_index_for_debug = 1;
char acm_local_for_debug = 0;
do {
if (acm_local_for_debug == '$') exit(0);
if (test_index_for_debug > 20)
throw runtime_error("Check the stdin!!!");
auto start_clock_for_debug = clock();
solve();
auto end_clock_for_debug = clock();
cout << "Test " << test_index_for_debug << " successful" << endl;
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "--------------------------------------------------" << endl;
} while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#else
solve();
#endif
return 0;
}