pku 3070 Fibonacci 矩阵快速幂相乘求Fibonacci 数列

http://poj.org/problem?id=3070

思路：

这里n很大单纯的递推是O(N)会超时，所以要用矩阵快速幂优化；

f(n) 1 1 f(1)

= 的n-1次方 *

f(n-1) 1 0 f(0)

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 10000
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 10000007
#define M 100007
#define N 5
using namespace std;
//freopen("din.txt","r",stdin);
struct Mat{
ll mat[3][3];
void init(){
for (int i = 0; i < 2; ++i){
for (int j = 0; j < 2; ++j){
mat[i][j] = 1;
}
}
mat[1][1] = 0;
}
};
Mat operator * (Mat a,Mat b){
Mat c;
int i,j,k;
CL(c.mat,0);
//矩阵乘积
for (i = 0; i < 2; ++i){
for (j = 0; j < 2; ++j){
for (k = 0; k < 2; ++k){
if (!a.mat[i][k] || !b.mat[k][j]) continue;
c.mat[i][j] += a.mat[i][k]*b.mat[k][j];
if (c.mat[i][j] > MOD) c.mat[i][j] %= MOD;
}
}
}
return c;
}
Mat operator ^ (Mat a,ll k){
Mat c;
int i,j;
//单位矩阵
for (i = 0; i < 2; ++i){
for (j = 0; j < 2; ++j){
c.mat[i][j] = (i == j);
}
}
//求k次幂
while (k){
if (k&1) c = c*a;
k >>= 1;
a = a*a;
}
return c;
}

int main(){
ll n;
Mat a;
while (~scanf("%I64d",&n)){
if (n == -1) break;
if (n == 0) {printf("0\n"); continue;}
a.init();//初始化
a = (a^(n - 1));//求n - 1次幂的到{fn,fn - 1};
printf("%I64d\n",a.mat[0][0]);
}
return 0;
}