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  • poj3070矩阵快速幂

    Fibonacci
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 7752   Accepted: 5501

    Description

    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

     1 #include <iostream>
     2 #include <stdio.h>
     3 #include <string.h>
     4 #include <math.h>
     5 #include <stdlib.h>
     6 
     7 using namespace std;
     8 int N;
     9 struct matrix
    10 {
    11        int a[2][2];
    12 }origin,res;
    13 matrix multiply(matrix x,matrix y)
    14 {
    15        matrix temp;
    16        memset(temp.a,0,sizeof(temp.a));
    17        for(int i=0;i<N;i++)
    18        {
    19                for(int j=0;j<N;j++)
    20                {
    21                        for(int k=0;k<N;k++)
    22                        {
    23                                temp.a[i][j]+=x.a[i][k]*y.a[k][j]%10000;
    24                                temp.a[i][j]%=10000;
    25                        }
    26                }
    27        }
    28        return temp;
    29 }
    30 void calc(int n)
    31 {
    32      memset(res.a,0,sizeof(res.a));
    33       origin.a[0][0]=1;origin.a[0][1]=1;
    34       origin.a[1][0]=1;origin.a[1][1]=0;
    35      for(int i=0;i<N;i++)
    36      res.a[i][i]=1;
    37      while(n)
    38      {
    39              if(n&1)
    40                     res=multiply(res,origin);
    41              n>>=1;
    42              origin=multiply(origin,origin);
    43      }
    44 }
    45 int main()
    46 {
    47     N=2;
    48     int n;
    49     while(cin>>n)
    50     {
    51         if(n==-1)
    52         break;
    53         if(n)
    54         calc(n-1);
    55         if(n)
    56         cout<<res.a[0][0]<<endl;
    57         else cout<<0<<endl;
    58     }
    59 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ERKE/p/3272728.html
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