洛谷P4593 [TJOI2018]教科书般的亵渎 【数学】

题目链接

洛谷P4593

题解

这种神仙数学题我当然不会做。。

orz dalao

不过推导倒是挺简单

由题我们只需模拟出代价，只需使用\(S(n,k) = \sum\limits_{i = 1}^{n} i^{k}\)这样的前缀和计算

我不知道怎么来的这样一个公式【如果有那位dalao知道欢迎留言】：

\[(n + 1)^{k + 1} - n^{k + 1} = \sum\limits_{i = 1}^{k + 1} {k + 1\choose i}n^{k + 1 - i}\]

我们发现这个公式右侧涵盖了所有\(n^i\)的项，如果我们将所有\(n\)枚举出来，将会的得到：

\[
\begin{aligned}
(n + 1)^{k + 1} - n^{k + 1} &= \sum\limits_{i = 1}^{k + 1} {k + 1\choose i}n^{k + 1 - i} \n^{k + 1} - (n - 1)^{k + 1} &= \sum\limits_{i = 1}^{k + 1} {k + 1\choose i}(n - 1)^{k + 1 - i} \......... \2^{k + 1} - 1^{k + 1} &= \sum\limits_{i = 1}^{k + 1} {k + 1\choose i}1^{k + 1 - i} \\end{aligned}
\]

全部相加，得到：

\[(n + 1)^{k + 1} - 1 = \sum\limits_{i= 1}^{k + 1} {k + 1 \choose i} S(n,k + 1 - i)\]

取出\(S(n,k)\)

\[S(n,k) = \frac{(n + 1)^{k + 1} - 1 - \sum\limits_{i = 2}^{k + 1}{k + 1 \choose i} S(n,k + 1 - i)}{k + 1}\]

发现就可以\(O(k^2)\)递推了

由于模拟也是\(O(k^2)\)的

所以最终复杂度\(O(k^4)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 105,maxm = 100005,INF = 1000000000,P = 1000000007;
inline LL read(){
    LL out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int n,m,K,fac[maxn],fv[maxn],inv[maxn];
LL a[maxn];
void init(){
    fac[0] = fac[1] = fv[0] = fv[1] = inv[0] = inv[1] = 1;
    for (int i = 2; i < maxn; i++){
        fac[i] = 1ll * fac[i - 1] * i % P;
        inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
        fv[i] = 1ll * fv[i - 1] * inv[i] % P;
    }
}
LL C(LL n,LL m){
    return 1ll * fac[n] * fv[m] % P * fv[n - m] % P;
}
LL qpow(LL a,LL b){
    LL ans = 1; a %= P;
    for (; b; b >>= 1,a = a * a % P)
        if (b & 1) ans = ans * a % P;
    return ans;
}
LL f[maxn];
LL S(LL n,LL k){
    if (!n) return 0;
    f[0] = n;
    for (int i = 1; i <= k; i++){
        LL tmp = 0;
        for (int j = 2; j <= i + 1; j++)
            tmp = (tmp + C(i + 1,j) * f[i + 1 - j]) % P;
        f[i] = (((qpow(n + 1,i + 1) - 1) % P - tmp) % P + P) % P * inv[i + 1] % P;
    }
    return f[k];
}
LL b[maxn];
int main(){
    init();
    int T = read();
    while (T--){
        n = read(); m = read(); K = m + 1;
        REP(i,m) a[i] = read(); a[K] = n + 1;
        sort(a + 1,a + 1 + K);
        LL ans = 0;
        for (int i = 0; i <= m; i++){
            for (int j = i; j <= m; j++){
                ans = ((ans + (S(a[j + 1] - 1,K) - S(a[j],K)) % P) % P + P) % P;
            }
            LL len = a[i + 1] - a[i];
            for (int j = i + 1; j <= K; j++) a[j] -= len;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/Mychael/p/9052067.html