题解

设前\(n\)个人的礼物个数为\(F_n\), 那么显然\[F_n = 2 \times F_{n-1} + i^k\]
考虑矩阵快速幂
棘手的问题是：\(i^k\)不是可以直接用矩阵乘法可以递推的东西
由二项式定理可得：\[a^k = \sum_{i = 1}^{k}(a-1)^i {k \choose i}\]
那么我们可以给\(\left( (i-1)^0\; (i-1)^1\;(i-1)^2\; \cdots\; (i-1)^k\;\right)\) 乘上一个杨辉三角矩阵， 就能得到\(\left(i^0\;i^1\;i^2\;\cdots\;i^k\right)\)
然后我们就能用矩阵快速幂算\(F\)， 显然\(Ans = F_{n-1} + i^k\)
于是就做完了

注意：这题两个n相乘可能会long long溢出， 因此算\(n^k\)前要先给\(n\)取模

代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>using namespace std;typedef long long LL;const LL mod = 1e9 + 7;LL power(LL a, LL n, LL mod)
{LL Ans = 1;while (n){if (n & 1) Ans = Ans * a % mod;a = a * a % mod;n >>= 1;}return Ans;
}LL n; int K;struct Matrix
{LL a[12][12];Matrix() { memset(a, 0, sizeof(a)); }LL* operator [] (int x) { return a[x]; }friend Matrix operator * (Matrix a, Matrix b){Matrix Ans;for (int i = 0; i <= K + 1; i++)for (int k = 0; k <= K + 1; k++)if (a[i][k])for (int j = 0; j <= K + 1; j++)(Ans[i][j] += a[i][k] * b[k][j] % mod) %= mod;return Ans;}
};Matrix power(Matrix a, LL n)
{Matrix Ans;for (int i = 0; i <= K + 1; i++) Ans[i][i] = 1;while (n){if (n & 1) Ans = Ans * a;a = a * a;n >>= 1;}return Ans;
}Matrix Ans, a;int main()
{Matrix A, Ans;scanf("%lld %d", &n, &K);if (n <= 2){LL Ans = 1;for (int i = 2; i < n; i++)Ans = (2 * Ans + power(i, K, mod)) % mod;Ans = (Ans + power(n, K, mod)) % mod;printf("%lld\n", Ans);return 0;}A[0][0] = 1;for (int i = 1; i <= K; i++){A[i][0] = 1; A[i][i] = 1;for (int j = 1; j < i; j++)A[i][j] = (A[i-1][j-1] + A[i-1][j]) % mod;}A[K+1][K+1] = 2;for (int i = 0; i <= K; i++)A[K+1][i] = A[K][i];for (int i = 0; i <= K; i++)Ans[i][1] = 1;Ans[K+1][1] = 1;Ans = power(A, n - 2) * Ans;printf("%lld\n", (Ans[K+1][1] + power(n%mod, K, mod)) % mod);return 0;
}