矩阵快速幂EOJ EOJ Monthly 2021.9 Sponsored by TuSimple A. Amazing Discovery
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Problem A: Amazing Discovery
评测传送门
蒟蒻赛时就开了这么一个题,推了好久也还是推不出来😣
看了人家官方题解后,顺利AC。但是,这个推导过程有点搞怪啊,不是看了题解,我还真推不出来这个递推关系。
官方题解:
题解传送门
下面我提供一些推导过程中的式子,可自行代入验证一波。
这里提供一个可以进行数学计算的超强👉代数计算工具网站
S
1
=
2
a
S_1 =2a
S1=2a
S
2
=
2
a
2
+
2
b
S_2 = 2a^2+2b
S2=2a2+2b
S
3
=
2
a
3
+
6
a
b
S_3 = 2a^3+6ab
S3=2a3+6ab
S
4
=
2
a
4
+
2
b
2
+
12
a
2
b
S_4 = 2a^4+2b^2+12a^2b
S4=2a4+2b2+12a2b
S
5
=
2
a
5
+
10
a
b
2
+
20
a
3
b
S_5 = 2a^5+10ab^2+20a^3b
S5=2a5+10ab2+20a3b
S
5
=
2
a
∗
S
4
−
(
a
2
−
b
)
S
3
S_5 = 2a * S_4 - (a^2-b)S_3
S5=2a∗S4−(a2−b)S3
矩阵快速幂中A矩阵的构造:
[
S
n
,
S
n
+
1
]
[S_n,S_n+1]
[Sn,Sn+1] ×
A
A
A =
[
S
n
+
1
,
S
n
+
2
]
[S_n+1,S_n+2]
[Sn+1,Sn+2]
A
=
[
0
b
−
a
2
1
2
a
]
A= \\begingathered \\beginbmatrix 0 & b-a^2 \\\\ 1 & 2a \\endbmatrix \\endgathered
A=[01b−a22a]
初始矩阵为:
F
0
=
[
2
,
2
a
]
F0 = [2,2a]
F0=[2,2a]
之后直接矩阵快速幂计算即可。
AcCoding:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int mod = 998244353;
const int N = 2;
/*
2a * S[n-1] - S[n] = (a - b * b) * S[n - 2]
f0[2] = S[n],S[n+1]
A =
0,2a,
1,b * b - a
*/
void mul(ll c[][N], ll a[][N], ll b[][N])
static ll tmp[N][N];
memset(tmp, 0, sizeof tmp);
for (int i = 0;i < N;i++)
for (int j = 0;j < N;j++)
for (int k = 0;k < N;k++)
(tmp[i][j] += a[i][k] % mod * b[k][j] % mod) %= mod;
memcpy(c, tmp, sizeof tmp);
int main()
ll a, b, n; scanf("%lld%lld%lld", &a, &b, &n);
ll f[N][N] = 2 * a, 2ll * a * a + 2ll* b ;//n = 1
ll A[N][N] =
0,((b - a * a) % mod + mod) % mod,
1,2ll * a % mod
;
n--;
while (n)
if (n & 1) mul(f, f, A);
mul(A, A, A);
n >>= 1;
ll res = (f[0][0] % mod + mod) % mod;
printf("%lld", res);
return 0;
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