POJ 3070 Fibonacci
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6065 | Accepted: 4246 |
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:
.
Source
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; void js(int (*a)[2],int (*c)[2]) { int b[2][2]; b[0][0]=(a[0][0]*c[0][0]+a[0][1]*c[1][0])%10000; b[0][1]=(a[0][0]*c[0][1]+a[0][1]*c[1][1])%10000; b[1][0]=(a[1][0]*c[0][0]+a[1][1]*c[1][0])%10000; b[1][1]=(a[1][0]*c[0][1]+a[1][1]*c[1][1])%10000; a[0][0]=b[0][0]; a[0][1]=b[0][1]; a[1][0]=b[1][0]; a[1][1]=b[1][1]; } int main() { int a[2][2],b[2][2]; int n; while(scanf("%d",&n),n>=0) { if(n==0) {printf("0\n"); continue;} a[0][0]=1; a[0][1]=1; a[1][0]=1; a[1][1]=0; b[0][0]=1; b[0][1]=0; b[1][0]=0; b[1][1]=1; for(;n>0;n=n>>1,js(a,a))//短小精悍呀、呵呵 if(n&1) js(b,a); printf("%d\n",b[0][1]); } return 0; }