「POJ3070」Fibonacci
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
|
1 2 3 4 5 |
0 9 999999999 1000000000 -1 |
Sample Output
|
1 2 3 4 |
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:
.
矩阵乘法快速幂入门。。
题目把题解说的很清楚了
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 |
#include<iostream> #include<cstdio> using namespace std; 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]; } int main() { while(scanf("%d",&n)) { if(n==-1)return 0; 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; while(n) { if(n&1)mul(a,b,b); n>>=1; mul(a,a,a); } printf("%d\n",b[1][0]); } } |
OI 课件=>
