HZAU 1202 GCD （矩阵快速幂 + GCD）

Problem Description

Xiao Ming found the compute time of gcd(fibn,fibn+1) is the most when he learnt the gcd, and the result of it is always fib1

He wants to know what gcd(1+Sn,1+Sm)

And gcd is greatest common divisor,

fib1=1,fib2=1,fibn=fibn−1+fibn−2(n≥3)

Sn=∑i=1nfibi

Input Description

The first line is an positive integer T. (1 ≤ T ≤ 10^3) indicates the number of test cases. In the next T lines, there are three positive integer n, m, p(1 ≤ n, m, p ≤ 10^9) at each line.

Output Description

In each test case, output the compute result of gcd(1+Sn,1+Sm)%p

Sample Input:

1
1 2 3

Sample Output:

1

题意

给出 n,m,p 三个整数，求斐波那契数列前 n 项和与前 m 项和的最大公约数模 p

思路

斐波那契数列有这样两条性质：

gcd(Fn,Fm)=Fgcd(n,m)

F1+F2+F3+F4+F5+...+Fn+1=Fn+2

于是，题目的答案便是 Fgcd(n,m)%p

最大公约数可以用辗转相除法求得，然后再利用矩阵快速幂得到斐波那契数第 k 项，记得模 p

AC 代码

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
using namespace std;
typedef long long LL;

LL n,m,mod;
LL gcd(LL a,LL b)
{
    if(b==0)return a;
    return gcd(b,a%b);
}

struct node
{
    LL mp[2][2];
    void init(LL a,LL b,LL c,LL d)
    {
        mp[0][0]=a;
        mp[0][1]=b;
        mp[1][0]=c;
        mp[1][1]=d;
    }
    void mult(node x,node y)                //两矩阵乘法
    {
        memset(mp,0,sizeof(mp));
        for(LL i=0; i<2; i++)
            for(LL j=0; j<2; j++)
                for(LL k=0; k<2; k++)
                    mp[i][j]=(mp[i][j]+x.mp[i][k]*y.mp[k][j])%mod;
    };
} init;

struct node expo(struct node x, LL k)  //进行k次幂运算
{
    struct node tmp;
    tmp.init(1,0,0,1);                      //单位矩阵
    while(k)                                //快速幂部分
    {
        if(k&1)tmp.mult(tmp,x);
        x.mult(x,x);
        k>>=1;
    }
    return tmp;
}

int main()
{
    int T;
    ios::sync_with_stdio(false);
    cin>>T;
    while(T--)
    {
        cin>>n>>m>>mod;
        int k=gcd(n+2,m+2);
        init.init(1,1,1,0);
        cout<<expo(init,k).mp[0][1]%mod<<endl;
    }
    return 0;
}