A Boring Problem

题解

其实这题不难,只要想到了前缀和差分就基本OK了.

我们要求的是第$i$项的式子:

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

记$S_i=a_1+a_2+...+a_i,S_0=0$

$F(i)=(S_i-S_0{)}^k+(S_i-S_1{)}^k+...+(S_i-S_{i-1}{)}^k$

二项式定理展开:

$F(i)={\sum }_{t=0}^k{C}_k^t{S}_i^t(-S_0{)}^{k-t}+{\sum }_{t=0}^k{C}_k^t{S}_i^t(-S_1{)}^{k-t}+...+{\sum }_{t=0}^k{C}_k^t{S}_i^t(-S_{i-1}{)}^{k-t}$

整理得:

$F(i)={\sum }_{t=0}^k{C}_k^t{S}_i^t(-1{)}^{k-t}(S_0^{k-t}+S_1^{k-t}+...+S_{i-1}^{k-t})$

再记

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

那么

$F(i)={\sum }_{t=0}^k{C}_k^t{S}_i^t(-1{)}^{k-t}(SS[k-t][i-1])$

注意到$SS$可以$O(nk)$预处理出来,$S$可以$O(n)$预处理出来,而$F(i)$就可以$O(k)$得到.

注意!$S_0^0=1$

代码

#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;
}