[bzoj3157][bzoj3516]國王奇遇記【數學】

【題目鏈接】
　　https://www.lydsy.com/JudgeOnline/problem.php?id=3157
【題解】
　　一道數學題，考慮使用擾動法。記f(k)=∑ni=1ik∗mi$f(k)=\sum _{i=1}^n{i}^k\ast m^i$
　　那麼有(k>0)$(k>0)$ ：m∗f(k)−f(k)=∑ni=1ik∗mi+1−∑ni=1ik∗mi$m\ast f(k)-f(k)=\sum _{i=1}^n{i}^k\ast m^{i+1}-\sum _{i=1}^n{i}^k\ast m^i$
(m−1)f(k)=∑n+1i=2(i−1)k∗mi−∑ni=1ik∗mi$(m-1)f(k)=\sum _{i=2}^{n+1}(i-1{)}^k\ast m^i-\sum _{i=1}^n{i}^k\ast m^i$
(m−1)f(k)=nk∗mn+1+∑ni=1(i−1)k∗mi−∑ni=1ik∗mi$(m-1)f(k)=n^k\ast m^{n+1}+\sum _{i=1}^n(i-1{)}^k\ast m^i-\sum _{i=1}^n{i}^k\ast m^i$
(m−1)f(k)=nk∗mn+1+∑ni=1mi∗((i−1)k−ik)$(m-1)f(k)=n^k\ast m^{n+1}+\sum _{i=1}^n{m}^i\ast ((i-1{)}^k-i^k)$
(m−1)f(k)=nk∗mn+1+∑ni=1mi∑k−1j=0C(k,j)∗(−1)k−j∗ij$(m-1)f(k)=n^k\ast m^{n+1}+\sum _{i=1}^n{m}^i\sum _{j=0}^{k-1}C(k,j)\ast (-1{)}^{k-j}\ast i^j$
(m−1)f(k)=nk∗mn+1+∑k−1j=0C(k,j)∗(−1)k−j∑ni=1ij∗mi$(m-1)f(k)=n^k\ast m^{n+1}+\sum _{j=0}^{k-1}C(k,j)\ast (-1{)}^{k-j}\sum _{i=1}^n{i}^j\ast m^i$
(m−1)f(k)=nk∗mn+1+∑k−1j=0C(k,j)∗(−1)k−jf(j)$(m-1)f(k)=n^k\ast m^{n+1}+\sum _{j=0}^{k-1}C(k,j)\ast (-1{)}^{k-j}f(j)$
f(k)=nk∗mn+1+∑k−1j=0C(k,j)∗(−1)k−jf(j)m−1$f(k)=\frac{n^k\ast m^{n+1}+\sum _{j=0}^{k-1}C(k,j)\ast (-1{)}^{k-j}f(j)}{m-1}$
f(m)$f(m)$ 就是所求的答案。接下來就可以遞推了。
時間複雜度O(M2)$O(M^2)$
【代碼】

/* - - - - - - - - - - - - - - -
    User :      VanishD
    problem :   [bzoj3157] 
    Points :    maths
- - - - - - - - - - - - - - - */
# include <bits/stdc++.h>
# define    ll      long long
# define    inf     0x3f3f3f3f
# define    M       2000
using namespace std;
int read(){
    int tmp = 0, fh = 1; char ch = getchar();
    while (ch < '0' || ch > '9'){ if (ch == '-') fh = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9'){ tmp = tmp * 10 + ch - '0'; ch = getchar(); }
    return tmp * fh;
}
const int P = 1e9 + 7;
int n, m, f[M + 1], c[M + 1][M + 1];
int power(int x, int y){
    int i = x; x = 1;
    while (y > 0){
        if (y % 2 == 1) x = 1ll * x * i % P;
        i = 1ll * i * i % P;
        y /= 2;
    }
    return x;
}
int pow2(int x){
    return (x % 2 == 1) ? -1 : 1;
}
int main(){
    n = read(), m = read();
    if (m == 1){
        printf("%lld\n", (1ll + n) * n / 2 % P);
        return 0;
    }
    c[0][0] = 1;
    for (int i = 1; i <= M; i++){
        c[i][0] = 1;
        for (int j = 1; j <= i; j++)
            c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % P;
    }
    f[0] = 1ll * (power(m, n + 1) - m) * power(m - 1, P - 2) % P;
    for (int i = 1; i <= m; i++){
        f[i] = 1ll * power(n, i) * power(m, n + 1) % P;
        for (int j = 0; j < i; j++)
            f[i] = (f[i] + 1ll * c[i][j] * pow2(i - j) * f[j]) % P;
        f[i] = 1ll * f[i] * power(m - 1, P - 2) % P;
    }
    printf("%d\n", (f[m] + P) % P);
    return 0;
}