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

//找周期 
#include <stdio.h>

#define N 20000
int a[N]={0,1};

int init()
{
     int i,j,k;
    for(i=2;i<N;i++)
    {
        a[i]=(a[i-1]+a[i-2])%10000;
        if(a[i]==0)
        {
            a[i+1]=(a[i]+a[i-1])%10000;
            if(a[i+1]==1)
                break;
        }
    }
    return i;
}

int main()
{
     int i,j,k;
     int n;
    int period = init();
    while(scanf("%d",&n),n!=-1)
    {
          n%=period;
        printf("%d\n",a[n]);
    }
    return  0;
}        

//快速幂算法 
 #include<stdio.h>
 
 void fast_pow(int a1[][2], int a2[][2]) 
 {
     int c[2][2];
     int i, j, k;
     for (i=0; i<2; ++i) 
      {
         for (j=0; j<2; ++j) 
           {
             c[i][j] = 0;
             for (k=0; k<2; ++k) 
                {
                 c[i][j] = (c[i][j] + a1[i][k] * a2[k][j]) % 10000;
             }
         }
     }
     for (i=0; i<2; ++i) 
      {
         for (j=0; j<2; ++j) 
           {
             a1[i][j] = c[i][j];
         }
     }
 }
 
 int main() 
 {
      int i,j,k;
     int n;
     while (scanf("%d", &n) , n != -1) 
     {
         int a[2][2] = {{1, 1}, {1, 0}};
         int b[2][2] = {{1, 0}, {0, 1}};//单位矩阵 
         while (n) 
         {
             if (n & 1) 
                 fast_pow(b, a);
             fast_pow(a, a);
             n >>= 1;
         }
         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);
     }
     return 0;
 }

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

 

#include<stdio.h>
int main()
{
    int f[200],a,b,n,i;
    while(scanf("%d%d%d",&a,&b,&n),a||b||n)
    {
        if(n>2)
        {
            f[1]=f[2]=1;
            for(i=3;i<200;i++)
            {
                f[i]=(a*f[i-1]+b*f[i-2])%7;
                if(f[i-1]==1&&f[i]==1)
                    break;
            }
            i-=2;//LPP，最小正周期
            n=n%i;//比如1~5是个循环周期，f(6)=f(1)、
            if(n==0)
                printf("%d\n",f[i]);
            else
                printf("%d\n",f[n]);
        }
        else
            printf("1\n");
    }
    return 0;
}