NYOJ 148

 

fibonacci数列(二)

时间限制： 1000 ms | 内存限制： 65535 KB

难度： 3

 

描述

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

0
9
1000000000
-1

样例输出

0
34
6875

 1 //找周期 
 2 #include <stdio.h>
 3 
 4 #define N 20000
 5 int a[N]={0,1};
 6 
 7 int init()
 8 {
 9      int i,j,k;
10     for(i=2;i<N;i++)
11     {
12         a[i]=(a[i-1]+a[i-2])%10000;
13         if(a[i]==0)
14         {
15             a[i+1]=(a[i]+a[i-1])%10000;
16             if(a[i+1]==1)
17                 break;
18         }
19     }
20     return i;
21 }
22 
23 int main()
24 {
25      int i,j,k;
26      int n;
27     int period = init();
28     while(scanf("%d",&n),n!=-1)
29     {
30           n%=period;
31         printf("%d\n",a[n]);
32     }
33     return  0;
34 }        

 

 1 //快速幂算法 
 2  #include<stdio.h>
 3  
 4  void fast_pow(int a1[][2], int a2[][2]) 
 5  {
 6      int c[2][2];
 7      int i, j, k;
 8      for (i=0; i<2; ++i) 
 9       {
10          for (j=0; j<2; ++j) 
11            {
12              c[i][j] = 0;
13              for (k=0; k<2; ++k) 
14                 {
15                  c[i][j] = (c[i][j] + a1[i][k] * a2[k][j]) % 10000;
16              }
17          }
18      }
19      for (i=0; i<2; ++i) 
20       {
21          for (j=0; j<2; ++j) 
22            {
23              a1[i][j] = c[i][j];
24          }
25      }
26  }
27  
28  int main() 
29  {
30       int i,j,k;
31      int n;
32      while (scanf("%d", &n) , n != -1) 
33      {
34          int a[2][2] = {{1, 1}, {1, 0}};
35          int b[2][2] = {{1, 0}, {0, 1}};//单位矩阵 
36          while (n) 
37          {
38              if (n & 1) 
39                  fast_pow(b, a);
40              fast_pow(a, a);
41              n >>= 1;
42          }
43          printf ("%d\n", b[1][0]);//最后n必然从1变为0，所以最后一次总要执行 fast_pow(b, a);
44      }
45      return 0;
46  }

HDU 1005

f(1) = 1, f(2) = 1, f(n) = (A * f(n – 1) + B * f(n – 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

 

同样是构造矩阵

 

 

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int f[200],a,b,n,i;
 5     while(scanf("%d%d%d",&a,&b,&n),a||b||n)
 6     {
 7         if(n>2)
 8         {
 9             f[1]=f[2]=1;
10             for(i=3;i<200;i++)
11             {
12                 f[i]=(a*f[i-1]+b*f[i-2])%7;
13                 if(f[i-1]==1&&f[i]==1)
14                     break;
15             }
16             i-=2;//LPP，最小正周期
17             n=n%i;//比如1~5是个循环周期，f(6)=f(1)、
18             if(n==0)
19                 printf("%d\n",f[i]);
20             else
21                 printf("%d\n",f[n]);
22         }
23         else
24             printf("1\n");
25     }
26     return 0;
27 }