题目链接
https://odzkskevi.qnssl.com/d474b5dd1cebae1d617e6c48f5aca598?v=1524578553
题意
给出一个表达式 算法 f(n)
思路
n 很大 自然想到是 矩阵快速幂
那么问题就是 怎么构造矩阵
我们想到的一种构造方法是
n = 2 时
n = 3 时
然后大概就能够发现规律了吧 。。
AC代码
#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <list>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>
#define CLR(a, b) memset(a, (b), sizeof(a))
#define pb push_back
#define bug puts("***bug***");
#define fi first
#define se second
#define stack_expand #pragma comment(linker, "/STACK:102400000,102400000")
#define syn_close ios::sync_with_stdio(false);cin.tie(0);
#define sp system("pause");
//#define bug
//#define gets gets_s
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <string, int> psi;
typedef pair <string, string> pss;
typedef pair <double, int> pdi;
const double PI = acos(-1.0);
const double E = exp(1.0);
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const int maxn = 1e2 + 10;
const int MOD = 142857;
int d, n, m;
ll a[20], b[20];
struct Matrix
{
ll a[20][20];
Matrix() {}
Matrix operator * (Matrix const &b)const
{
Matrix res;
CLR(res.a, 0);
for (int i = 0; i < d; i++)
for (int j = 0; j < d; j++)
for (int k = 0; k < d; k++)
res.a[i][j] = (res.a[i][j] + this->a[i][k] * b.a[k][j]) % m;
return res;
}
};
Matrix pow_mod(Matrix ans, int n)
{
Matrix base;
CLR(base.a, 0);
for (int i = 0; i < d; ++i)
{
base.a[i][0] = a[i];
}
for (int i = 0; i < d; ++i)
{
base.a[i][i + 1] = 1;
}
while (n > 0)
{
if (n & 1)
ans = ans * base;
base = base * base;
n >>= 1;
}
return ans;
}
int main()
{
while (scanf("%d %d %d", &d, &n, &m) && (d || n || m))
{
for (int i = 0; i < d; i++)
scanf("%lld", &a[i]);
for (int i = 0; i < d; i++)
scanf("%lld", &b[i]);
if (n <= d)
{
printf("%lld
", b[n - 1] % m);
continue;
}
Matrix ans;
for (int i = 0; i < d; i++)
for (int j = 0; j < d; j++)
ans.a[i][j] = b[d - j - 1];
ans = pow_mod(ans, n - d);
printf("%lld
", ans.a[0][0]);
}
return 0;
}