遞推

ssl 1532

題目大意

給出數(shù)列$a_0,a_1\dots a_n$和$f$的前$n?1$項(xiàng)$f_0,f_1\dots f_{n?1}$
$f_i=a_0?f_{i?n}+a_1?f_{i?(n?1)}+...+a_{n?1}?f_{i?1}+a_n$
現(xiàn)在讓你求$f_k$(結(jié)果對(duì)9973取模）

輸入樣例

2 10 1 1 0 0 1

輸出樣例

55

數(shù)據(jù)范圍

$$\begin{gathered} 1?n?k?10^{18} \\ 1?a_i,fi?10^4 \end{gathered}$$

解題思路

k較大，無法直接遞推，我們考慮矩陣乘法
設(shè)矩陣
$\left[\begin{array}{ccccc}{f}_{i?n}& f_{i?(n?1)}& \dots & f_{i?1}& 1\end{array}\right]$
$f_{i?n}?f_{i?2}$直接等于下一位
$f_{i?1}$就是前面n項(xiàng)乘對(duì)應(yīng)的a，然后加上最后一個(gè)1乘$a_n$
1不變
得出需要乘的矩陣后快速冪即可

代碼

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long #define wyc 9973 ll n, k; struct matrix {ll n, m, a[100][100];matrix operator *(matrix const &b)const{matrix c;c.n = n;c.m = b.m;for (int i = 1; i <= c.n; ++i)for (int j = 1; j <= c.m; ++j)c.a[i][j] = 0;for (int i = 1; i <= c.n; ++i)for (int k = 1; k <= m; ++k)for (int j = 1; j <= c.m; ++j)c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j] % wyc) % wyc;return c;} }A, B; void Counting(ll g) {while(g){if (g&1) A = A * B;B = B * B;g>>=1;} } int main() {scanf("%lld%lld", &n, &k);B.n = B.m = A.m = n + 1;A.n = 1;for (int i = 1; i <= n + 1; ++i){scanf("%lld", &B.a[i][n]);//前n個(gè)數(shù)乘上對(duì)應(yīng)的aif (i < n) B.a[i + 1][i] = 1;//等于下一個(gè)}B.a[n + 1][n + 1] = 1;//最后一個(gè)afor (int i = 1; i <= n; ++i)scanf("%lld", &A.a[1][i]);A.a[1][n + 1] = 1;Counting(k);printf("%lld", A.a[1][1]);return 0; }

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