洛谷- P1306 斐波那契公约数

P1306 斐波那契公约数：https://www.luogu.org/problemnew/show/P1306

这道题目就是求第n项和第m项的斐波那契数字，然后让这两个数求GCD，输出答案的后8位；

思路：

　　1/gcd(F[n],F[m])=F[gcd(n,m)]。所以先求出gcd（n，m），然后构造斐波那契数列的矩阵快速幂。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;       
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e8;
const double esp = 1e-8;
const double PI=acos(-1.0);



template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/
            
            struct mat{
                int r,c;
                ll a[4][4];
            };

            int gcd(int a,int b){
                if(b == 0)return a;
                return gcd(b, a%b);
            }
            mat mul(mat a,mat b){
                mat tmp = a;
                tmp.a[0][0] = tmp.a[0][1] = tmp.a[1][0] = tmp.a[1][1] = 0;
                for(int i=0; i< a.r; i++){
                    for(int j=0; j<b.c;j++){
                        for(int k=0; k<a.c; k++){
                            tmp.a[i][j] += a.a[i][k] * b.a[k][j];
                            tmp.a[i][j] %= mod;
                        }
                    }
                }
                return tmp;
            }

            int ksm(int n){
                mat ans,ic;
                ic.r = ic.c = 2;
                ic.a[0][0] = ic.a[0][1] = ic.a[1][0] = 1;
                ic.a[1][1] = 0;

                ans.r = 1,ans.c = 2;
                ans.a[0][0] = ans.a[0][1] = 1;
                while(n > 0){
                    if(n&1) ans = mul(ans, ic);
                    ic = mul(ic,ic);
                    n>>=1;
                }
                return ans.a[0][0];
            }
int main(){
            int a,b;
            scanf("%d%d", &a, &b);
            int n = gcd(a, b);
            if(n <=2)cout<<1<<endl;
            else cout<<ksm(n-2)<<endl;
            return 0;
}