QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#461868#2438. Minimum Spanning Treesalpha1022AC ✓54ms4056kbC++143.4kb2024-07-03 08:42:212024-07-03 08:42:21

Judging History

你现在查看的是最新测评结果

  • [2024-07-03 08:42:21]
  • 评测
  • 测评结果:AC
  • 用时:54ms
  • 内存:4056kb
  • [2024-07-03 08:42:21]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

using ll = long long;

const int mod = 1e9 + 7;

void add(int &x, int y) { if ((x += y - mod) < 0) x += mod; }
void sub(int &x, int y) { if ((x -= y) < 0) x += mod; }
int mpow(int a, int b) {
  int ret = 1;
  for (; b; b >>= 1) {
    if (b & 1) ret = (ll)ret * a % mod;
    a = (ll)a * a % mod;
  }
  return ret;
}

const int K = 4;
const int N = 40;
const int S = K * N;
const int LIM = N * (N - 1) / 2;

int n, k, tk;
int p[K + 5], fac[N + 5], ifac[N + 5], inv[N + 5];
int pw[LIM + 5], ipw[LIM + 5];
int f[K + 5][N + 5][S + 5], g[N + 5][S + 5], h[N + 5][S + 5];
int ans[S + 5];

void solve() {
  scanf("%d%d", &n, &k); int nv = mpow(100, mod - 2);
  for (int i = 0; i <= k; ++i) scanf("%d", p + i), p[i] = (ll)p[i] * nv % mod;
  for (tk = k; !p[k]; --k);
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = (ll)fac[i - 1] * i % mod;
  ifac[n] = mpow(fac[n], mod - 2);
  for (int i = n; i; --i) ifac[i - 1] = (ll)ifac[i] * i % mod;
  for (int i = 1; i <= n; ++i) inv[i] = (ll)ifac[i] * fac[i - 1] % mod;
  memset(f, 0, sizeof f);
  f[0][1][0] = 1;
  for (int v = 1, q = 1; v <= k; ++v) {
    int iq = mpow(q, mod - 2);
    pw[0] = ipw[0] = 1;
    for (int i = 1; i <= n * (n - 1) / 2; ++i)
      pw[i] = (ll)pw[i - 1] * q % mod,
      ipw[i] = (ll)ipw[i - 1] * iq % mod;
    memset(h, 0, sizeof h);
    for (int i = 1; i <= n; ++i)
      for (int s = 0; s <= (v - 1) * (i - 1); ++s)
        if (f[v - 1][i][s])
          h[i][s + v] = (ll)f[v - 1][i][s] * ipw[i * (i - 1) / 2] % mod;
    memset(g, 0, sizeof g), g[0][0] = 1;
    for (int i = 1; i <= n; ++i)
      for (int s = 0; s <= v * i; ++s) {
        for (int j = 1; j <= i; ++j)
          for (int t = 0; t <= (v - 1) * j + 1 && t <= s; ++t)
            if (h[j][t] && g[i - j][s - t])
              g[i][s] = (g[i][s] + (ll)j * h[j][t] % mod * g[i - j][s - t]) % mod;
        g[i][s] = (ll)g[i][s] * inv[i] % mod;
      }
    for (int i = 0; i <= n; ++i)
      for (int s = 0; s <= v * i; ++s)
        g[i][s] = (ll)g[i][s] * pw[i * (i - 1) / 2] % mod;
    if (p[0] == 0 && v == k) {
      for (int i = 1; i <= n; ++i)
        for (int s = v; s <= v * i; ++s)
          f[v][i][s - v] = g[i][s];
      break;
    }
    sub(q, p[v]), iq = mpow(q, mod - 2);
    for (int i = 1; i <= n * (n - 1) / 2; ++i)
      pw[i] = (ll)pw[i - 1] * q % mod,
      ipw[i] = (ll)ipw[i - 1] * iq % mod;
    for (int i = 0; i <= n; ++i)
      for (int s = 0; s <= v * i; ++s)
        g[i][s] = (ll)g[i][s] * ipw[i * (i - 1) / 2] % mod;
    memset(h, 0, sizeof h);
    for (int i = 1; i <= n; ++i)
      for (int s = 0; s <= v * i; ++s) {
        h[i][s] = (ll)g[i][s] * i % mod;
        for (int j = 1; j < i; ++j)
          for (int t = 0; t <= v * j && t <= s; ++t)
            if (h[j][t] && g[i - j][s - t])
              h[i][s] = (h[i][s] + (ll)(mod - j) * h[j][t] % mod * g[i - j][s - t]) % mod;
        h[i][s] = (ll)h[i][s] * inv[i] % mod;
      }
    for (int i = 1; i <= n; ++i)
      for (int s = v; s <= v * i; ++s)
        f[v][i][s - v] = (ll)h[i][s] * pw[i * (i - 1) / 2] % mod;
  }
  for (int i = 0; i <= k * (n - 1); ++i) ans[i] = (ll)f[k][n][i] * fac[n] % mod;
  fill(ans + k * (n - 1) + 1, ans + tk * (n - 1) + 1, 0);
  k = tk;
  for (int i = n - 1; i <= k * (n - 1); ++i) printf("%d%c", ans[i], " \n"[i == k * (n - 1)]);
}

int T;

int main() {
  for (scanf("%d", &T); T; --T) solve();
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 54ms
memory: 4056kb

input:

200
3 1
50 50
3 2
0 50 50
3 3
25 25 25 25
8 1
41 59
7 3
37 30 7 26
3 3
16 12 18 54
9 2
9 43 48
9 3
3 40 42 15
9 1
29 71
9 2
40 42 18
5 1
76 24
5 1
39 61
9 2
23 38 39
10 4
18 15 34 2 31
7 2
23 28 49
9 4
15 13 25 19 28
7 1
64 36
6 1
50 50
9 1
4 96
4 1
64 36
9 2
24 45 31
9 2
3 61 36
9 1
65 35
8 4
6 1 3...

output:

500000004
500000004 375000003 125000001
406250003 109375001 250000002 265625002 562500004
858129220
40267248 73443306 307645653 13908396 542571454 781149891 223877799 478284083 469782292 483514097 271207900 851118600 686534546
708608005 271088002 536992004 107032001 243224002
763536836 20527108 7248...

result:

ok 200 lines