洛谷-1306

洛谷-1306

思路

引理1：（$f_n$是斐波那契数列）

$$gcd(f_n,f_{n+1})=1$$

引理2：（设$m>n$）

$$gcd(f_n,f_m)=gcd(f_n,f_{m-n}\ast f_{n+1})$$

结论：

$$gcd(f_n,f_m)=f_{gcd(n,m)}$$

证明：这个

注意：

斐波那契数列是$f_1=1,f_2=1,f_3=2$

而不是$f_0=1,f_1=1,f_2=2$

Code

#include <bits/stdc++.h>
using namespace std;
#define _u_u_ ios::sync_with_stdio(false), cin.tie(nullptr)
#define cf int _o_o_;cin>>_o_o_;for (int Case = 1; Case <= _o_o_;Case++)
#define SZ(x) (int)(x.size())
inline void _A_A_();
signed main() {_A_A_();return 0;}

using ll = long long;
#define int ll
int mod = 1e8;
const int maxn = 2e5 + 10;
const int N = 2, M = 5010;
const int inf = 0x3f3f3f3f;

struct mat
{
    int m[N][N];
};

mat operator * (const mat&a, const mat&b) {
    mat c;
    memset(c.m, 0, sizeof c.m);
    for (int i = 0;i < N;i++) {
        for (int j = 0;j < N;j++) {
            for (int k = 0;k < N;k++) {
                c.m[i][j] = (c.m[i][j] + a.m[i][k] * b.m[k][j]) % mod;
            }
        }
    }
    return c;
}

mat qpow(mat a, int n) {
    mat c;
    memset(c.m, 0, sizeof c.m);
    for (int i = 0;i < N;i++) c.m[i][i] = 1;
    while (n ) {
        if (n & 1) {
            c = c * a;
        }
        a = a * a;
        n >>= 1;
    }
    return c;
}

inline void _A_A_() {
    #ifdef LOCAL
    freopen("in.in", "r", stdin);
    #endif
    int n, m;
    cin >> n >> m;
    mat s;
    memset(s.m,0,sizeof s.m);
    s.m[0][0] = s.m[0][1] = s.m[1][0] = 1;
    s = qpow(s, __gcd(n,m) - 1);
    n = (s.m[0][1] + s.m[1][1]) % mod;
    cout << n << "\n";
}