ProjectEuler237 Tours on a 4 x n playing board
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思路是这样的
插头dp-->打表-->OEIS查表-->通项公式-->矩阵快速幂优化线性递推
OEIS竟然有这个东西的生成函数啊
答案为15836928
这是最终代码
#include <bits/stdc++.h>
using namespace std;
#define MOD 191981
#define mod 100000000
#define ll long long
struct Matrix {
vector<vector<int> > a;
int n, m;
void init(int n, int m) {
this->n = n, this->m = m;
a.resize(n);
for(int i = 0; i < n; ++i) a[i].resize(m);
}
vector<int>& operator [](int index_x) {
return a[index_x];
}
friend Matrix operator * (Matrix lhs, Matrix rhs) {
Matrix c;
c.init(lhs.n, rhs.m);
for(int i = 0; i < lhs.n; ++i)
for(int j = 0; j < rhs.m; ++j)
for(int k = 0; k < lhs.m; ++k)
c[i][j] = (c[i][j]+1LL*lhs[i][k]*rhs[k][j]%mod)%mod;
return c;
}
};
Matrix eye(int n) {
Matrix c;
c.init(n, n);
for(int i = 0; i < n; ++i)
c[i][i] = 1;
return c;
}
Matrix fpow(Matrix x, ll p) {
Matrix ret = eye(x.n);
while(p) {
if(p&1) ret = ret*x;
x = x*x;
p >>= 1;
}
return ret;
}
//a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) + a(n-4)
int main() {
ll n;
scanf("%lld", &n);
if(n == 1) printf("1\n");
else if(n == 2) printf("1\n");
else if(n == 3) printf("4\n");
else if(n == 4) printf("8\n");
else {
Matrix m0, p, ans;
m0.init(4, 1);
m0[0][0] = 8, m0[1][0] = 4, m0[2][0] = 1, m0[3][0] = 1;
p.init(4, 4);
p[0][0] = 2, p[0][1] = 2, p[0][2] = -2, p[0][3] = p[1][0] = p[2][1] = p[3][2] = 1;
ans = fpow(p, n-4)*m0;
printf("%d\n", ans[0][0]);
}
return 0;
}
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