A Boring Problem

題解

其實這題不難,只要想到了字首和差分就基本OK了.

我們要求的是第$i$項的式子:

$F(i)=(a_1+a_2+...+a_i)^k+(a2+...+a_i)^k+...+(a_i)^k$

記$S_i = a_1+a_2+...+a_i,S_0=0$

$F\left(i\right)=(S_i-S_0{)}^k+(S_i-S_1{)}^k+\mathrm{.}\mathrm{.}\mathrm{.}+(S_{}$

二項式定理展開:

$F(i)=\sum_{t=0}^kC_k^tS_i^t(-S_0)^{k-t}+\sum_{t=0}^kC_k^tS_i^t(-S_1)^{k-t}+...+\sum_{t=0}^kC_k^tS_i^t(-S_{i-1})^{k-t}$

整理得:

$F(i) = \sum_{t=0}^kC_k^tS_i^t(-1)^{k-t}(S_0^{k-t}+S_1^{k-t}+...+S_{i-1}^{k-t})$

再記

$SS[i][j] =S_0^i+S_1^i+...+S_{j-1}^i$

那麼

$F(i) = \sum_{t=0}^kC_k^tS_i^t(-1)^{k-t}(SS[k-t][i-1])$

注意到$SS$可以$O(nk)$預處理出來,$S$可以$O(n)$預處理出來,而$F(i)$就可以$O(k)$得到.

注意!$S_0^0 = 1$

程式碼

#include <iostream>
#include <algorithm>
#include <cstring>
#define pr(x) std::cout << #x << ':' << x << std::endl
#define rep(i,a,b) for(int i = a;i <= b;++i)

typedef long long LL;
const int N = 100010;
const LL P = 1e9+7;
int T,n,k;
char s[N];
long long S[101][N],SS[101][N];
long long C[110][110];
void init() {
	C[0][0] = 1;
	for(int i = 1;i <= 100;++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;
		}
	}
}
int main() {
	std::ios::sync_with_stdio(false);
	init();
	std::cin >> T;
	while(T--) {
		std::cin >> n >> k;
		std::cin >> s;
		for(int i = 0;i <= n;++i) S[0][i] = 1;
		for(int i = 1;i <= n;++i) S[1][i] = (s[i-1]-'0') + S[1][i-1] ;
		for(int i = 2;i <= k;++i) 
			for(int j = 1;j <= n;++j)
				S[i][j] = S[1][j] * S[i-1][j] % P;
		
		SS[0][0] = 1;

		for(int i = 0;i <= k;++i) {
			for(int j = 1;j <= n;++j) 
				SS[i][j] = (SS[i][j-1] + S[i][j])% P;
		}
		
		for(int i = 1;i <= n;++i) {
			long long ans = 0;
			for(int j = 0;j <= k;++j) {
				long long res = C[k][j]*S[j][i]%P*SS[k-j][i-1]%P;
				if((k-j)%2==0) ans = (ans + res) % P;
				else ans = (ans - res + P) % P;
			}
			if(i != 1) std::cout << " ";
			std::cout << ans;
		}
		std::cout << std::endl;
	}
	return 0;
}