poj3070 Fibonacci(矩阵快速幂)

矩阵的快速幂算法

/*
#include<iostream>
using namespace std;
typedef unsigned long long ll;
#define MOD 10000
struct Mat
{

};
ll f[2];
ll fib(int n)
{ //int f[2];
f[1]=0;f[2]=1;
for(int i=2;i<=n;i++)
{
f[0]=f[1]%MOD;
f[1]=f[2]%MOD;
f[2]=(f[1]+f[0])%MOD;
}
return f[2];
}
int main()
{ int n;
while(cin>>n&&n!=-1)
{ if(n==0)
cout<<0<<endl;
else if(n==1)
cout<<1<<endl;
else
cout<<fib(n)%MOD<<endl;
}

return 0;
}*/

#include <cstdio>
#include <iostream>
#include<cstdlib>
using namespace std;
typedef unsigned long long ll;
const int MOD = 10000;
/*
struct matrix
{
int m[2][2];
}ans, base;

matrix multi(matrix a, matrix b)
{
matrix tmp;
for(int i = 0; i < 2; ++i)
{
for(int j = 0; j < 2; ++j)
{
tmp.m[i][j] = 0;
for(int k = 0; k < 2; ++k)
tmp.m[i][j] = (tmp.m[i][j]+ a.m[i][k] * b.m[k][j])%MOD ;
}
}
return tmp;
}
int fast_mod(int n) // 求矩阵 base 的 n 次幂
{
base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;
base.m[1][1] = 0;
ans.m[0][0] = ans.m[1][1] = 1; // ans 初始化为单位矩阵
ans.m[0][1] = ans.m[1][0] = 0;
while(n)
{
if(n & 1) //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t
{
ans = multi(ans, base);
}
base = multi(base, base);
n >>= 1;
}
return ans.m[0][1];
}*/
struct Mat { ll mat[2][2];};
Mat operator * (Mat a, Mat b) {
Mat c;
memset(c.mat, 0, sizeof(c.mat));
int i, j, k;
for(k = 0; k < 2; ++k) {
for(i = 0; i < 2; ++i) {
if(a.mat[i][k] <= 0) continue;
for(j = 0; j < 2; ++j) {
if(b.mat[k][j] <= 0) continue;
c.mat[i][j] =( c.mat[i][j]+a.mat[i][k] * b.mat[k][j])%MOD;
}
}
}
return c;
}
Mat operator ^ (Mat a, int k)
{
Mat c;
int i,j;
for(i=0;i<2;i++)
for(j=0;j<2;j++)
c.mat[i][j]=(i==j);
for(;k;k>>=1)
{ if(k&1) c=c*a;
a=a*a;

}
return c;
}
int main()
{ Mat A,B;
A.mat[0][0]=1;
A.mat[0][1]=1;
A.mat[1][0]=1;
A.mat[1][1]=0;
int n;
while(cin>>n && n != -1)
{ B=A^n;
cout<<B.mat[1][0]%MOD<<endl;
}
return 0;
}