矩阵 Fibonacci

Fibonacci

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 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 F_n.

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 F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

099999999991000000000-1

Sample Output

0346266875

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:

.

 

矩阵相乘+快速幂

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define ll __int64#define mod 10000using namespace std;struct mat{    int p[2][2];};mat fun(mat a, mat b){ //矩阵相乘    int i, j, k;    mat c;    for(i=0; i<2; i++){        for(j=0; j<2; j++){            c.p[i][j] = 0;            for(k=0; k<2; k++){                c.p[i][j] += (a.p[i][k]%mod)*(b.p[k][j]%mod);                c.p[i][j] % mod;            }        }    }    return c;}mat quick(mat a, ll n){ //快速幂 a^n    mat b;    int i, j, k;    for(i=0; i<2; i++){        for(j=0; j<2; j++){            if(i==j)                b.p[i][j] = 1;            else                b.p[i][j] = 0;        }    }    while(n){        if(n%2==1)            b = fun(b,a);        a = fun(a, a);        n /= 2;    }    return b;}int main(){    ll n;    mat a;    a.p[0][0] = 1;    a.p[0][1] = 1;    a.p[1][0] = 1;    a.p[1][1] = 0;    while(cin>>n){        if(n==-1)            break;        mat x = quick(a,n); //a^n        cout<<x.p[0][1]%mod<<endl;    }    return 0;}