POJ - 3070 - Fibonacci ( 矩阵快速幂 )

描述

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the
Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is


.

Given an integer n, your goal is to compute the last 4 digits of Fn.

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by


.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:


.

输入The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
输出For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
样例输入

0
9
1000000000
-1

样例输出

0
34
6875

#include<cstdio>
#include<cstring>
#include<iostream>
#define MOD 10000
using namespace std;
int n;
struct Matrix{
int matrix[2][2];
}m1,m2,m3;
//矩阵乘法
Matrix f1(Matrix m1,Matrix m2){
Matrix m3;
memset(m3.matrix,0,sizeof(m3.matrix));
for(int i=0 ;i<2;i++){
for(int j=0 ;j<2 ;j++){
for(int x=0 ;x<2 ;x++){
m3.matrix[i][j]+=(m1.matrix[i][x]*m2.matrix[x][j])%MOD;
}
}
}
return m3;
}

//快速幂
Matrix f2(Matrix m4,int n){
Matrix m5;
if(n==1){
return m4;
}
else if(n==2){
return f1(m4,m4);
}
else if(n%2==0){
m5 = f2(f1(m4,m4),n/2);
return m5;
}
else{
m5 = f2(f1(m4,m4),n/2);
return f1(m5,m4);
}
}

int main(){
while(cin>>n){
if(n==0)cout<<0<<endl;
else if(n==-1)break;
else{
Matrix ans;
Matrix temp;
Matrix in;
in.matrix[0][0]=1;
in.matrix[0][1]=1;
in.matrix[1][0]=1;
in.matrix[1][1]=0;
ans = f2(in,n);
cout<<ans.matrix[0][1]%MOD<<endl;
}

}

return 0;
}

#include<cstdio>
#include<cstring>
#define MOD 10000
#define N 31
using namespace std;
int n;
struct mat{
long long v

;
mat(){memset(v,0,sizeof(v));}
};

mat mul(mat x,mat y){
mat ans;
for(int i= 0;i<n ;i++){
for(int j=0 ;j<n ;j++){
for(int k=0 ;k<n ;k++){
ans.v[i][j] = (ans.v[i][j]+x.v[i][k]*y.v[k][j])%MOD;
}
}
}
return ans;

}
//乘法
mat pow(mat x ,int k){
mat ans;
//初始化ans为单位矩阵
ans.v[0][0] = 1 ; ans.v[0][1] = 0;
ans.v[1][0] = 0; ans.v[1][1] = 1;
while(k){
if(k&1) ans = mul(ans,x) ;
x = mul(x,x);
k >>= 1;
}
return ans;
}

int main(){
int k;
n=2;
mat m1 ;
m1.v[0][0] = 1 ; m1.v[0][1] = 1;
m1.v[1][0] = 1; m1.v[1][1] = 0;
while(scanf("%d",&k)!=EOF&&k!=-1){
mat ans;
ans = pow(m1,k);
printf("%lld\n",ans.v[0][1]);
}

return 0;
}