【POJ】3070 - Fibonacci（矩阵快速幂）

最新推荐文章于 2019-03-28 17:11:09 发布

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最新推荐文章于 2019-03-28 17:11:09 发布

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分类专栏：矩阵

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本文介绍了一种高效计算斐波那契数列中任意项最后四位数字的方法，使用矩阵快速幂运算来减少时间复杂度，适用于大整数运算场景。

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Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 12951 Accepted: 9210
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

Stanford Local 2006
入门题目
代码

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXN =100+10;
const int mod=10000;
struct Matrix{
    LL a[MAXN][MAXN];
    int r,c;
};
Matrix ori,res;
void init(){
    memset(res.a,0,sizeof(res.a));
    res.r=2;res.c=2;
    for(int i=1;i<=2;i++)  //将结果矩阵赋值为单位矩阵
        res.a[i][i]=1;//单位矩阵A就是 [未知矩阵]B*A=B;
    ori.r=ori.c=2;
    ori.a[1][1]=ori.a[1][2]=ori.a[2][1]=1;// 初始化初始矩阵
    ori.a[2][2]=0;
}
Matrix multi(Matrix x,Matrix y){
    Matrix z;
    memset(z.a,0,sizeof(z.a));
    z.r=x.r;  z.c=y.c; //新矩阵的行数等于x的行数，列数等于y的列数
    for(int i=1;i<=x.r;i++){
        for(int k=1;k<=x.c;k++){
            if(x.a[i][k]==0) continue;
            for(int j=1;j<=y.c;j++)
                z.a[i][j]=(z.a[i][j]+(x.a[i][k]*y.a[k][j]%mod))%mod;
        }
    }
    return z;
}
void Matrix_mod(int n){
    while(n){
        if(n&1) res=multi(ori,res);
        ori=multi(ori,ori);
        n>>=1;
    }
    printf("%lld\n",res.a[1][2]%mod);
}
int main(){
    int n;
    while(scanf("%d",&n)&&(n!=-1)){
        init();
        Matrix_mod(n);
    }
    return 0;
}