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\n", 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\n", ans.a[0][0]); } return 0; }转载于:https://www.cnblogs.com/Dup4/p/9433073.html
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