poj3070 矩阵快速幂Fib

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, andF_n =F_{n − 1} +F_{n − 2} forn ≥ 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.

Input

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.

Output

For each test case, print the last four digits of F_n. If the last four digits ofF_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., printF_n mod 10000).

Sample Input

0 9 999999999 1000000000 -1

Sample Output

0 34 626 6875

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:

.

思路：照着模板学；

#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> typedef long long LL; using namespace std; const int MOD=10000; struct mat { int m[2][2]; }ans; //计算A*B mat mul(mat a,mat b) { mat c; for(int i=0;i<2;i++) //左行，新矩阵行 for(int j=0;j<2;j++) //右列 ，新矩阵列 { c.m[i][j]=0; for(int k=0;k<2;k++) //左列右行 { c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%MOD; } } return c; } //计算A^n int pow(mat a,int n) { memset(ans.m,0,sizeof(ans.m)); //初始化为单位矩阵 for(int i=0;i<2;i++) ans.m[i][i]=1; while(n) //二分思想，快速幂 { if(n&1) ans=mul(ans,a); a=mul(a,a); n>>=1; } return ans.m[1][0]; //或者ans.m[0][1] } int main() { int n; mat base={1,1,1,0}; while(scanf("%d",&n) && n!=-1) printf("%d\n",pow(base,n)); return 0; }

vector 建立二维数组来写：

#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<queue> #define max_n 10010 typedef long long LL; using namespace std; const int MOD=10000; typedef vector<int> vec; typedef vector<vec> mat; //计算A*B mat mul(mat &A,mat &B) { mat C(A.size(),vec(B[0].size())); for(int i=0;i<A.size();i++) for(int k=0;k<B.size();k++) for(int j=0;j<B[0].size();j++) C[i][j]=(C[i][j]+A[i][k]*B[k][j])%MOD; return C; } //计算A^n mat pow(mat A,int n) { mat B(A.size(),vec(A.size())); for(int i=0;i<A.size();i++) B[i][i]=1; while(n) { if(n&1) B=mul(B,A); A=mul(A,A); n>>=1; } return B; } int n; void solve() { mat A(2,vec(2)); A[0][0]=1;A[0][1]=1; A[1][0]=1;A[1][1]=0; A=pow(A,n); printf("%d\n",A[1][0]); } int main() { while(scanf("%d",&n) && n!=-1) { solve(); } return 0; }