講解:點擊打開鏈接
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 15917 | Accepted: 11188 |
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
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:
.
題目描述
斐波那契數列是由如下遞推式定義的數列
F0=0
F1=1
Fn+1=Fn+1+Fn
求這個數列第n項的值對104取餘後的結果。
限制條件
⋅ 0⩽n⩽1016
我也是剛剛纔搞懂了矩陣乘法(如果你不知道什麼是矩陣乘法的話,右轉百度百科),於是來應用一下新知識,如有表述不到位的地方請見諒。
下面進入正文
首先,我們先介紹一下對於斐波那契數列如何求解。把斐波那契數列的遞推式表示成矩陣就得到下面的式子
(Fn+2Fn+1)= (1110)(Fn+1Fn)
我們發現式子裏有個固定的矩陣 (1110)
記這個矩陣爲A,則有
(Fn+1Fn)=An(F1F0)=An(10)
因此只要求出An就可以求出Fn了。關於An的計算我們可以採用類似快速冪的算法,在O(logn)時間裏求出第n項的值。(轉自:點擊打開鏈接)
#include<cstdio>
int n,a[2][2],b[2][2];
void mul(int a[2][2],int b[2][2],int ans[2][2])
{
int t[2][2];
for (int i=0;i<2;i++)
for (int j=0;j<2;j++)
{
t[i][j]=0;
for (int k=0;k<2;k++)
t[i][j]=(t[i][j]+a[i][k]*b[k][j])%10000;
}
for (int i=0;i<2;i++)
for (int j=0;j<2;j++)
ans[i][j]=t[i][j];
}
void pow(int k)
{
while (k)
{
if (k%2==1)
mul(a,b,b);
k/=2;
mul(a,a,a);
}
}
int main()
{
while (scanf("%d",&n)!=EOF)
{
if (n==-1)
break;
a[0][0]=a[0][1]=a[1][0]=1;a[1][1]=0;
b[0][0]=b[1][1]=1;
b[1][0]=b[0][1]=0;
pow(n);
printf("%d\n",b[1][0]);
}
return 0;
}
這道題還碰見一種神奇的操作,我還沒理解到底是什麼規律。。
來自:(點擊打開鏈接)
#include<stdio.h>
int a[100050];
void f()
{
a[0] = 0;
a[1] = 1;
for (int i = 2; i <= 100050; i ++)
{
a[i] = (a[i - 1] + a[i - 2]) % 10000;
}
}
int main ()
{
int n;
while (scanf("%d", &n) != EOF)
{
if(n == -1) break;
f();
printf("%d\n", a[n % 15000]);
}
return 0;
}