2020 BIT冬训-二分三分快速幂矩阵 H - Fibonacci POJ - 3070 （矩阵快速幂）

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

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:

.

这题是矩阵快速幂。用二维数组表示矩阵，然后temp[i][j]+=(a[i][k]*d[k][j])来更新矩阵。（矩阵相乘还是麻烦的……）

需要特判的是0和1的时候直接返回值即可。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,ans=1,d[2][2],a[2][2],temp[2][2];
int main(){
    d[0][0]=1,d[1][0]=1,d[0][1]=1,a[0][0]=1,a[1][1]=1;
    while(scanf("%d",&n)&&n!=-1){
        if(n==0){
            printf("0\n");
            continue;
        }else if(n==1){
            printf("1\n");
            continue;
        }
        memset(d,0,sizeof(d));
        memset(a,0,sizeof(a));
        d[0][0]=1,d[1][0]=1,d[0][1]=1,a[0][0]=1,a[1][1]=1;
        while(n){
            if(n&1){
                memset(temp,0,sizeof(temp));
                for(int i=0;i<2;i++)
                    for(int j=0;j<2;j++)
                        for(int k=0;k<2;k++)
                            temp[i][j]+=(d[i][k]*a[k][j])%10000;
                for(int i=0;i<2;i++)
                    for(int j=0;j<2;j++)
                        a[i][j]=temp[i][j]%10000;    
            }
            memset(temp,0,sizeof(temp));
            for(int i=0;i<2;i++)
                    for(int j=0;j<2;j++)
                        for(int k=0;k<2;k++)
                            temp[i][j]+=(d[i][k]*d[k][j])%10000;
            for(int i=0;i<2;i++)
                    for(int j=0;j<2;j++)
                        d[i][j]=temp[i][j]%10000;    
            n>>=1;                
        }
        if(a[1][1]==0)
            printf("0000\n");
        else
            printf("%d\n",a[0][1]);
    }
}