UVA - 10229 - Modular Fibonacci （矩陣快速冪 + fibonacci）

標籤：acm   uva   fibonacci   矩陣快速冪   

題目傳送：UVA - 10229

思路：就是簡單的矩陣快速冪求fibonacci數列，然後注意可能中間結果會爆int，因為2^19有50多萬

AC代碼：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <deque>#include <cctype>#define LL long long#define INF 0x7fffffffusing namespace std;int mp[20];LL n, m;void init() {mp[0] = 1;for(int i = 1; i < 20; i ++) {mp[i] = 2 * mp[i - 1];}}struct matrix {LL m[2][2];};matrix ans;matrix base = {1, 1, 1, 0};matrix multiply(matrix a, matrix b, LL MOD) {matrix ret;for(int i = 0; i < 2; i ++) {for(int j = 0; j < 2; j ++) {ret.m[i][j] = 0;for(int k = 0; k < 2; k ++) {ret.m[i][j] = (ret.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;}}}return ret;}LL kmod(matrix a, LL n, LL MOD) {matrix ans = {1, 0, 0, 1};while(n) {if(n & 1) ans = multiply(ans, a, MOD);a = multiply(a, a, MOD);n >>= 1;}return ans.m[0][1];}int main() {init();while(cin >> n >> m) {cout << kmod(base, n, mp[m]) << endl;}return 0;}

UVA - 10229 - Modular Fibonacci （矩陣快速冪 +　fibonacci）