poj 3070 Fibonacci
http://poj.org/problem?id=3070
入门级矩阵快速幂,自己打了个属于自己的模板!
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <algorithm> 5 6 using namespace std; 7 typedef __int64 ll; 8 const int mod = 10000; 9 const int matSize = 100; 10 11 struct Matrix { 12 ll val[matSize][matSize]; 13 14 Matrix(bool Init = false) { 15 for (int i = 0; i < matSize; i++) { 16 for (int j = 0; j < matSize; j++) { 17 val[i][j] = 0; 18 } 19 if (Init) val[i][i] = 1; 20 } 21 } 22 23 24 void print() { 25 for (int i = 0; i < matSize; i++) { 26 for (int j = 0; j < matSize; j++) { 27 printf("%I64d ", val[i][j]); 28 } 29 puts(""); 30 } 31 puts("~~~"); 32 } 33 } Base; 34 35 Matrix operator * (Matrix _a, Matrix _b) { 36 Matrix ret = Matrix(); 37 38 for (int i = 0; i < matSize; i++) { 39 for (int k = 0; k < matSize; k++) { 40 if (_a.val[i][k]) { 41 for (int j = 0; j < matSize; j++) { 42 ret.val[i][j] += _a.val[i][k] * _b.val[k][j]; 43 ret.val[i][j] %= mod; 44 } 45 } 46 } 47 } 48 49 return ret; 50 } 51 52 Matrix operator ^ (Matrix _a, ll _p) { 53 Matrix ret = Matrix(true); 54 55 while (_p) { 56 if (_p & 1) { 57 ret = ret * _a; 58 } 59 _a = _a * _a; 60 _p >>= 1; 61 } 62 63 return ret; 64 } 65 66 void initBase(ll _k = 1, ll _a = 1) { 67 Base.val[0][1] = _k; 68 Base.val[1][1] = _a; 69 Base.val[0][0] = 0; 70 Base.val[1][0] = 1; 71 } 72 73 ll fun(ll _x) { 74 Matrix ret = Matrix(); 75 76 ret.val[0][0] = 0; 77 ret.val[0][1] = 1; 78 79 ret = ret * (Base ^ _x); 80 // ret.print(); 81 82 return ret.val[0][0]; 83 } 84 85 ll deal(ll _x) { 86 return fun(_x); 87 } 88 89 int main() { 90 ll n; 91 92 initBase(); 93 // Base.print(); 94 while (~scanf("%I64d", &n) && ~n) { 95 printf("%I64d\n", deal(n)); 96 } 97 98 return 0; 99 }
——wirtten by Lyon