POJ 3070 —— Fibonacci 【矩阵快速幂】

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Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12190 Accepted: 8651

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

#include <cstdio>#include <iostream>#include <vector>#include <assert.h>using namespace std;const int MOD = (int)1e4;struct Mat {    vector<vector<int> > m;    Mat() {}    Mat(int a, int b) {        m.resize(a);        for(int i=0; i<m.size(); i++) {            m[i].resize(b);        }    }};Mat operator * (const Mat& a, const Mat& b) {    assert(a.m[0].size() == b.m.size());    Mat c(a.m.size(), b.m[0].size());    for(int i=0; i<a.m.size(); i++) {        for(int j=0; j<b.m[0].size(); j++) {            c.m[i][j] = 0;            for(int k=0; k<b.m.size(); k++) {                c.m[i][j] = (c.m[i][j] + a.m[i][k] * b.m[k][j] % MOD) % MOD;            }        }    }    return c;}Mat operator ^ (Mat a, int k) {    assert(a.m.size() == a.m[0].size());    Mat c(a.m.size(), a.m.size());        for(int i=0; i<a.m.size(); i++) {        for(int j=0; j<a.m.size(); j++) {            c.m[i][j] = (i==j);        }    }        while(k) {        if(k&1) {            c = c * a;            }        a = a * a;        k>>=1;    }    return c;}int main (){    int n;    Mat a(2,2), b(2,1);    a.m[0][0]=a.m[1][0]=a.m[0][1]=1;    a.m[1][1]=0;        b.m[0][0]=1;    b.m[1][0]=0;        while(scanf("%d", &n) != EOF && n != -1) {        if(n==0)    printf("%d\n", b.m[1][0]);        else if(n==1)    printf("%d\n", b.m[0][0]);        else {            Mat ans = (a^(n-1)) * b;            printf("%d\n", ans.m[0][0]);        }    }        return 0;}

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