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 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 |
[Submit] [Go Back] [Status] [Discuss]
题意其实说的很清楚,斐波那契数列的第n项即为1 1的n次方的后的第一项,由于让输出后四位,即对10000取余即可,
1 0
#include<iostream> #include<algorithm> using namespace std; typedef long long ll; #define MAXN 105 const int mod=10000; struct mat { ll m[MAXN][MAXN]; }unit; ll n; mat msub(mat a,mat b) { mat ans; ll x=0; for(int i=0;i<2;i++) { for(int j=0;j<2;j++) { x=0; for(int k=0;k<2;k++) { x+=((a.m[i][k]*b.m[k][j])%mod); } ans.m[i][j]=x%mod; } } return ans; } mat qpow(mat a,ll x) { mat ret=unit; while(x) { if(x&1) ret=msub(ret,a); a=msub(a,a); x>>=1; } return ret; } void init_unit() { for(int i=0;i<MAXN;i++) { unit.m[i][i]=1; } } ll solve(ll n) { init_unit(); mat a; a.m[0][0]=1; a.m[0][1]=1; a.m[1][0]=1; a.m[1][1]=0; a=qpow(a,n); return a.m[0][1]; } int main() { while(cin>>n&&n!=-1) { cout<<solve(n)<<endl; } return 0; }