首页 > 代码库 > 快速Fibonacci数列,矩阵法
快速Fibonacci数列,矩阵法
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 9156 | Accepted: 6494 |
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:
.
这个题是要用矩阵的知识解决斐波那契数列,题目中给的公式很简单,直接往上套就行了,那我们就先看除了题目给的方法外的另一个方法,当然还是矩阵。
先看下面的公式:
不难看出这个式子是把:
写成矩阵的形式而已。
将上面的式子迭代可得:
故可得一般式为:
继续展开会得到:
设:
则:
所以:
现在已经很明显了,其中还要用到矩阵快速幂的知识,代码如下:
#include <stdio.h> #include <string.h> #include <math.h> typedef __int64 int64; int64 c[2][2],ans[2][2],d[2][2],a[2][2]; int main() { int64 i,j,n,k,t; while(scanf("%I64d",&n)!=EOF) { if(n==-1) break; if(n==0) { printf("0\n"); continue; } if(n==1) { printf("1\n"); continue; } memset(ans,0,sizeof(ans)); for(i=0;i<2;i++) ans[i][i]=1; a[0][0]=a[0][1]=a[1][0]=1; a[1][1]=0; n=n-2; while(n!=0) { if(n%2==1) { memset(d,0,sizeof(d)); for(i=0;i<2;i++) for(j=0;j<2;j++) if(a[i][j]) for(k=0;k<2;k++) { d[i][k]+=ans[i][j]*a[j][k]; d[i][k]=d[i][k]%10000; } for(i=0;i<2;i++) for(j=0;j<2;j++) ans[i][j]=d[i][j]; } memset(c,0,sizeof(c)); for(i=0;i<2;i++) for(j=0;j<2;j++) { if(a[i][j]==0) continue; for(k=0;k<2;k++) { c[i][k]+=a[i][j]*a[j][k]; c[i][k]=c[i][k]%10000; } } for(i=0;i<2;i++) for(j=0;j<2;j++) a[i][j]=c[i][j]; n=n/2; } t=(ans[0][0]+ans[0][1])%10000; if(t==0) printf("0\n"); else printf("%I64d\n",t); } return 0; }