- 239's solution
-
P239's Solution
- @ 2026-4-6 17:19:39
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;
}