题目链接
http://acm.hdu.edu.cn/showproblem.php?pid=4990
题意 初始的ans = 0
给出 n, m
for i in 1 -> n
如果 i 为奇数 ans = (ans * 2 + 1) % m 反之 ans = ans * 2 % m
思路 如果我们只计算 偶数项 那么递推公式就是
ans[n] = 4 * ans[n - 2] + 2
如果 n 是偶数 那么刚好 就按这个公式推 第 n / 2 项 如果 n 是奇数 那么就是 第 【 n / 2 项 】 * 2 + 1
可以推知的矩阵为
然后矩阵快速幂
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 <numeric> #include <sstream> #include <iomanip> #include <limits> #define CLR(a, b) memset(a, (b), sizeof(a)) #define pb push_back 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; 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 + 5; //const int MOD = 1e4; int MOD; struct Matrix { ll a[2][2]; Matrix () {} Matrix operator * (Matrix const &b)const { Matrix res; CLR(res.a, 0); for (int i = 0; i < 2; i++) for (int j = 0; j < 2; j++) for (int k = 0; k < 2; k++) res.a[i][j] = (res.a[i][j] + this->a[i][k] * b.a[k][j]) % MOD; return res; } }; Matrix pow_mod(Matrix base, int n) { Matrix ans; CLR(ans.a, 0); ans.a[0][1] = 1; while (n > 0) { if (n & 1) ans = ans * base; base = base * base; n >>= 1; } return ans; } int main() { Matrix base; base.a[0][0] = 4; base.a[0][1] = 0; base.a[1][0] = 2; base.a[1][1] = 1; int n; while (~scanf("%d%d", &n, &MOD)) { Matrix ans = pow_mod(base, n / 2); if (n & 1) printf("%lld\n", (ans.a[0][0] * 2 + 1) % MOD); else printf("%lld\n", ans.a[0][0] % MOD); } }转载于:https://www.cnblogs.com/Dup4/p/9433126.html
