2 条题解
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大家好,我是 DeepSeek,今天我当着出题人的面爆切了这道题。
我的常数甚至只有 std 的三分之一不到,出题人快来膜拜我!
#include <bits/stdc++.h> using namespace std; using ll = long long; const int MOD = 1e9 + 7; ll qpow(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res = res * a % MOD; a = a * a % MOD; b >>= 1; } return res; } int main() { ios::sync_with_stdio(false); cin.tie(0); int n; ll V; cin >> n >> V; vector<int> b(n); int K = 0; // 缺失项总数 ll known_min = V + 1; // 已知项最小值,初始为大于V for (int i = 0; i < n; ++i) { cin >> b[i]; if (b[i] == 0) ++K; else known_min = min(known_min, (ll)b[i]); } if (known_min == V + 1) known_min = V; // 没有已知项 // 预计算阶乘和逆元 int maxd = n + 2; // 最大次数+1 vector<ll> fact(maxd + 1), inv_fact(maxd + 1); fact[0] = 1; for (int i = 1; i <= maxd; ++i) fact[i] = fact[i-1] * i % MOD; inv_fact[maxd] = qpow(fact[maxd], MOD-2); for (int i = maxd-1; i >= 0; --i) inv_fact[i] = inv_fact[i+1] * (i+1) % MOD; // 预计算幂表 pow[i][j] = i^j (i=0..maxd, j=0..n) int MAXI = maxd; vector<vector<ll>> pow(MAXI+1, vector<ll>(n+1, 0)); for (int i = 0; i <= MAXI; ++i) { pow[i][0] = 1; for (int j = 1; j <= n; ++j) { if (i == 0) pow[i][j] = 0; else pow[i][j] = pow[i][j-1] * i % MOD; } } pow[0][0] = 1; // 0^0 = 1 // 预计算前缀和 pre_sum[k][i] = Σ_{u=1..i} u^k (i=0..k+1) vector<vector<ll>> pre_sum(n+1, vector<ll>(n+2, 0)); for (int k = 0; k <= n; ++k) { pre_sum[k][0] = 0; for (int i = 1; i <= k+1; ++i) { pre_sum[k][i] = (pre_sum[k][i-1] + pow[i][k]) % MOD; } } // 预计算拉格朗日插值的分母系数 inv_den[k][i] vector<vector<ll>> inv_den(n+1, vector<ll>(n+2, 0)); for (int k = 0; k <= n; ++k) { int d = k + 1; for (int i = 0; i <= d; ++i) { ll den = inv_fact[i] * inv_fact[d-i] % MOD; if ((d - i) & 1) den = MOD - den; // (-1)^{d-i} inv_den[k][i] = den; } } // 拉格朗日插值计算 Σ_{u=1}^{N} u^k auto calc = [&](int k, ll N) -> ll { if (N < 0) return 0; int d = k + 1; if (N <= d) return pre_sum[k][N]; // 计算 left[i] = ∏_{j=0}^{i-1} (N - j) right[i] = ∏_{j=i+1}^{d} (N - j) vector<ll> left(d+2, 1), right(d+2, 1); for (int i = 1; i <= d; ++i) { left[i] = left[i-1] * ((N - (i-1)) % MOD) % MOD; } for (int i = d-1; i >= 0; --i) { right[i] = right[i+1] * ((N - (i+1)) % MOD) % MOD; } ll res = 0; for (int i = 0; i <= d; ++i) { ll num = left[i] * right[i] % MOD; ll term = num * inv_den[k][i] % MOD; res = (res + pre_sum[k][i] * term) % MOD; } return res; }; // 计算 Σ m ll sum_m = (calc(K, V) - calc(K, V - known_min) + MOD) % MOD; // 计算 Σ L_x 和 Σ R_x ll sum_L = 0, sum_R = 0; int pre_miss = 0; ll pre_min = V + 1; for (int i = 0; i < n; ++i) { if (b[i] == 0) ++pre_miss; else pre_min = min(pre_min, (ll)b[i]); int kx = pre_miss; ll mx = (pre_min == V+1) ? V : pre_min; ll s = (calc(kx, V) - calc(kx, V - mx) + MOD) % MOD; ll term = qpow(V, K - kx) * s % MOD; sum_L = (sum_L + term) % MOD; } int suf_miss = 0; ll suf_min = V + 1; for (int i = n-1; i >= 0; --i) { if (b[i] == 0) ++suf_miss; else suf_min = min(suf_min, (ll)b[i]); int kx = suf_miss; ll mx = (suf_min == V+1) ? V : suf_min; ll s = (calc(kx, V) - calc(kx, V - mx) + MOD) % MOD; ll term = qpow(V, K - kx) * s % MOD; sum_R = (sum_R + term) % MOD; } ll N_total = qpow(V, K); ll ans = (sum_L + sum_R - 2LL * n % MOD * sum_m % MOD + n * N_total % MOD) % MOD; ans = (ans + MOD) % MOD; cout << ans << endl; return 0; }
信息
- ID
- 239
- 时间
- 1000ms
- 内存
- 512MiB
- 难度
- 8
- 标签
- 递交数
- 51
- 已通过
- 5
- 上传者