nyoj148fibonacci数列(二)

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

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

091000000000-1

样例输出

0346875

代码：

    #include <cstdio>      #include <iostream>      #include <vector>            using namespace std;      typedef vector<int> vec;      typedef vector<vec> mat;      typedef long long LL;      const int N = 10000;      mat mul(mat a,mat b)  //矩阵乘法      {          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] ) % N;              }          }          return c;      }            mat solve_pow(mat a,int n) //快速幂      {          mat b(a.size(),vec(a.size()));          for(int i=0;i<a.size();i++)              b[i][i]=1;          while(n>0)          {              if(n & 1)                  b=mul(b,a);              a=mul(a,a);              n >>= 1;          }                return b;      }      LL n;      void solve()      {          mat a(2,vec(2));          while(~scanf("%d",&n) && n!=-1)          {              a[0][0]=1,a[0][1]=1;              a[1][0]=1,a[1][1]=0;              a=solve_pow(a,n);              printf("%d\n",a[1][0]);          }      }      int main()      {          solve();          return 0;      }