[牛客练习赛69] D. 火柴排队多维dp+逆元+递推求排列组合-白红宇

[牛客练习赛69] D. 火柴排队多维dp+逆元+递推求排列组合

阅读量：334 次

发布时间：2019-03-04

本文共 1983 字，大约阅读时间需要 6 分钟。

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???????n???a?????????k??????????????d??????????????????????a_i < a_j???????a_i' < a_j'????

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??????????????????????????dp[i][j][k]??i??????j????????k???i?????????

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????????dp[i][j][0] = dp[i-1][j][0] + dp[i-1][j][1] * (a[i-1] + d ? a[i])

???????dp[i][j][1] = dp[i-1][j-1][0] + dp[i-1][j-1][1]

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                     using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const ll mod = 998244353;const int inf = 0x3f3f3f3f;const int maxn = 5000 + 10;void get_comb(int n) { c[0] = 1; for (int i = 1; i <= n; ++i) { c[i] = (n - i + 1) * c[i - 1] % mod * inv(i) % mod; }}ll inv(ll a) { return pow(a, mod - 2, mod);}void main() { ll n, d; scanf("%lld %lld", &n, &d); get_comb(n); vector
                     
                       a(n + 1); for (int i = 1; i <= n; ++i) { scanf("%lld", &a[i]); } sort(a + 1, a + n + 1); dp[1][1][1] = 1; dp[1][0][0] = 1; for (int i = 2; i <= n; ++i) { for (int j = 0; j <= i; ++j) { if ((i - 1) & 1) { dp[i & 1][j][0] = (dp[i - 1 & 1][j][0] + dp[i - 1 & 1][j][1] * (a[i - 1] + d <= a[i])) % mod; dp[i & 1][j][1] = (dp[i - 1 & 1][j - 1][1] + dp[i - 1 & 1][j - 1][0]) % mod; } else { dp[i & 1][j][0] = (dp[i - 1][j][0] + dp[i - 1][j][1] * (a[i - 1] + d <= a[i])) % mod; dp[i & 1][j][1] = (dp[i - 1][j - 1][1] + dp[i - 1][j - 1][0]) % mod; } } } for (int i = 1; i <= n; ++i) { ll ans = (dp[n & 1][i][0] + dp[n & 1][i][1]) % mod; ans = ans * pow(c[i], mod - 2, mod) % mod; ans = (ans + mod) % mod; printf("%lld\n", ans); }}

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????????dp[1][0][0]?dp[1][1][1]????1?

??????????????????????dp??

????????k????????????????????????

????????O(n^2)???????????????n=5000????

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