POJ - 3070 - Fibonacci ( 矩阵快速幂 )

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n ? 1} + F_{n ? 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 F_n.

Hint

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[N][N];	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;
}