北京区域赛I题,Uva7676,A Boring Problem,前缀和差分

转载自https://blog.csdn.net/weixin_37517391/article/details/83821752

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:

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

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：

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

整理得：

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

再记：

$SS[i][j] = S_0^i + S_1^i + ... + S_j^i$

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，$S$也可以$O(nk)$预处理出来，而$F(i)$就可以$O(k)$出来。

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;   //特殊化处理，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;
}