题目链接：http://poj.org/problem?id=3070
题目大意：
求第N项的Fibonacci数的后四位。
题目思路：
根据公式：
用矩阵快速幂就OK，模板题……但还是TLE了一次，原因是题目要求输入-1结束，我没看到……o(╯□╰)o
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 using namespace std; 7 const int MAX = 2; 8 const int M = 10000; 9 typedef struct { 10 int m[MAX][MAX]; 11 }Matrix; 12 Matrix a, per; 13 int n; 14 void init() { 15 int i , j; 16 for (i = 0; i < n; ++i) { 17 for (j = 0; j < n; ++j) { 18 per.m[i][j] = (i == j); 19 } 20 } 21 a.m[0][0] = a.m[0][1] = a.m[1][0] = 1; a.m[1][1] = 0; 22 } 23 Matrix multi(Matrix a, Matrix b) { 24 Matrix c; 25 int k, i, j; 26 for ( i = 0; i < n; ++i) { 27 for (j = 0; j < n; ++j) { 28 c.m[i][j] = 0; 29 for (k = 0; k < n; ++k) { 30 c.m[i][j] += a.m[i][k] * b.m[k][j]; 31 } 32 c.m[i][j] %= M; 33 } 34 } 35 return c; 36 } 37 Matrix power(int k) { 38 Matrix p, ans = per; 39 p = a; 40 while (k) { 41 if (k & 1) { 42 ans = multi(ans, p); k--; 43 } 44 else { 45 k /= 2; p = multi(p, p); 46 } 47 } 48 return ans; 49 } 50 51 int main(void) { 52 Matrix ans; int k; n = 2; 53 init(); 54 while (~scanf("%d", &k)) { 55 if (k == -1) break; 56 init(); 57 ans = power(k); 58 printf("%d\n", ans.m[1][0]); 59 } 60 return 0; 61 }
其实就是完完全全的模板……
