洛谷P4593 [TJOI2018]教科书般的亵渎 【数学】

题目链接

洛谷P4593

题解

orz dalao

upd:经典的自然数幂和，伯努利数裸题

由题我们只需模拟出代价，只需使用(S(n,k) = sumlimits_{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：

[(n + 1)^{k} - n^{k} = sumlimits_{i = 1}^{k} {k choose i}n^{k - i} ]

这玩意怎么来的呢？
左边为((n + 1)^k - n^k)，((n+1)^k)可以看做有(k)个位置进行染色，每个位置有(n + 1)种染色的方案数，减去(n^k)，就代表了拥有第(n + 1)种颜色的染色方案数
那么这个等式就很好理解了，我们枚举第(n + 1)种颜色染了多少个，就得到了右式

我们发现这个公式右侧涵盖了所有(n^i quad[ i in [0,k]])的项，我们令(k = k + 1)，如果我们将所有(n)枚举出来，将会的得到：

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

全部相加，得到：

[(n + 1)^{k + 1} - 1 = sumlimits_{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 - sumlimits_{i = 0}^{k - 1}{k + 1 choose i} S(n,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;
}
LL a[maxn],fac[maxn],fv[maxn],inv[maxn],n,m,K;
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 % P;
	for (int i = 1; i <= k; i++){
		LL tmp = 0;
		for (int j = 0; j <= i - 1; j++)
			tmp = (tmp + C(i + 1,j) * f[j] % P) % 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
",ans);
	}
	return 0;
}