Poj 3070 矩阵快速幂（找规律法）

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.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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:

.

这道题的意思就是让你输出第N个斐波那契数列的后四位是什么，不用保留前导零。这题的数据非常之大。就算O(n)的算法也会超时，当然这个题正确的解法是矩阵快速幂，不过这个题目比较特殊，通过时建坤大佬得知，结果是会循环然后通过测试得知结果是会循环的。
我们开一个非常大的数组，只保留后四位，即对1e4进行取余，这样并不会影响结果，按照加法的算法，低位次会对高位次产生影响，但高位次不会对低位次产生影响。当我们发现后面的数中如果又出现了这一位是1，前一位是0，即发生了循环。我得到他的循环长度。然后对循环长度取余数再从数组中输出就行了，完整代码如下：

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const int MAX = 1e6+10;const int mod = 1e4;int f[MAX];int Getans(){    f[0] = 0;f[1] = 1;    for(int i=2;i<=MAX-10;++i){        f[i] = (f[i-1] + f[i-2]) % mod;//在数组中只保留后四位。        if(f[i] == 1 && f[i-1] == 0){//又一次到达最开始的数，因此会循环            return (i-1);//从第i-1个开始循环，循环最后一个数是i-2，因为从0开始所以其循环长度是i-1,因此我们对i-1取余数。        }    }    return false;}int main(void){    int Len = Getans();    int n;    while(cin >> n && n!= -1){        cout << f[n%Len] << endl;    }    return 0;}

矩阵快速幂的做法如下：

#include<cstdio>#include<cstring>#include<cmath>#include<vector>#include<iostream>#include<algorithm>#include<set>using namespace std;const int mod = 10000;typedef vector<int> vec;typedef vector<vec> mat;typedef long long ll;mat mul(mat &A,mat &B){    mat C(A.size(),vec(B[0].size()));    for(int i=0;i<A.size();i++){        for(int j=0;j<B[0].size();j++){            for(int k = 0;k<B.size();k++){                C[i][j] = (C[i][j] + A[i][k]*B[k][j]) %mod;            }        }    }    return C;}mat pow(mat A,ll 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 main(void){    mat A(2,vec(2));//申请2个存放vec类型的mat，而且vec里面只有2个元素    ll n;    while(scanf("%lld",&n) && n!=-1){        A[0][0] = A[0][1] = A[1][0] = 1;        A[1][1] = 0;        A = pow(A,n);        printf("%d\n",A[1][0]);    }    return 0;}