luoguP4000 斐波那契数列

题目链接

luoguP4000 斐波那契数列

题解

根据这个东西

https://www.cnblogs.com/sssy/p/9418732.html

我们可以找出%p意义下的循环节

然后就可以做了

人傻，自带，大，常数

代码

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const LL maxn = 1000007;
LL dp[maxn * 10];
LL prime[maxn],s = 0;
bool vis[maxn];
void init_prime() {
    for(LL i = 2;i < maxn;i ++) {
        if(!vis[i]) prime[s ++] = i;
        for(LL j = 0;j < s && i * prime[j] < maxn;j ++) {
            vis[i * prime[j]] = 1;
            if(i%prime[j] == 0) break;
        }
    }
}
LL pow_mod(LL a1,LL b1){
    LL ret = 1;
    for(;b1;b1 >>= 1,a1 *= a1)
        if(b1 & 1) ret = ret * a1;
    return ret;
}
LL pow_mod2(LL a,LL b,LL p) {
    LL ret = 1;
    for(;b;b >>= 1,a = a * a % p)
        if(b & 1) ret = ret * a % p;
    return ret;
}
LL gcd(LL a,LL b) { return b ? gcd(b,a % b) : a; }
bool f(LL n,LL p) {
    if(pow_mod2(n,(p - 1) >> 1,p) != 1) return false;
    return true;
}
struct matrix{
    LL x1,x2,x3,x4;
};
matrix matrix_a,matrix_b;
matrix M2(matrix aa,matrix bb,LL mod){
    matrix tmp;
    tmp.x1 = (aa.x1 * bb.x1 % mod + aa.x2 * bb.x3 % mod) % mod;
    tmp.x2 = (aa.x1 * bb.x2 % mod + aa.x2 * bb.x4 % mod) % mod;
    tmp.x3 = (aa.x3 * bb.x1 % mod + aa.x4 * bb.x3 % mod) % mod;
    tmp.x4 = (aa.x3 * bb.x2 % mod + aa.x4 * bb.x4 % mod) % mod;
    return tmp;
}
matrix M(LL n,LL mod){
    matrix a,b;
    a=matrix_a;b=matrix_b;
    for(;n;n >>= 1,a = M2(a,a,mod))
        if(n & 1) b = M2(b,a,mod);
    return b;
}
LL fac[100][2],l,x,fs[1000];
void dfs(LL count,LL step) {
    if(step == l) {
        fs[x ++] = count;
        return ;
    }
    LL sum = 1;
    for(LL i = 0;i < fac[step][1];++ i) {
        sum *= fac[step][0];
        dfs(count * sum,step + 1);
    }
    dfs(count,step + 1);
}
LL solve2(LL p){
    if(p < 1e6 && dp[p])   return dp[p];
    bool ok = f(5,p);//判断5是否为p的二次剩余
    LL t;
    if(ok) t = p - 1;
    else t = 2 * p + 2;
    l = 0;
    for(LL i = 0;i < s;i ++){
        if(prime[i] > t / prime[i])  break;
        if(t % prime[i] == 0) {
            LL count = 0;
            fac[l][0] = prime[i];
            while(t % prime[i] == 0) count ++ , t /= prime[i];
            fac[l ++][1] = count;
        }
    }
    if(t > 1) fac[l][0]=t, fac[l++][1]=1;
    x = 0;
    dfs(1 , 0); //求t的因子
    sort(fs,fs + x);
    for(LL i = 0;i < x;i ++) {
        matrix m1 = M(fs[i],p);
        if(m1.x1 == m1.x4 && m1.x1 == 1 && m1.x2 == m1.x3 && m1.x2 == 0) {
            if(p < 1e6) dp[p] = fs[i];
            return fs[i];
        }
    }
}
LL get_M(LL n){
    LL ans = 1,cnt;
    for(LL i = 0;i < s;i ++) {
        if(prime[i] > n / prime[i]) break;
        if(n % prime[i] == 0) {
             LL count = 0;
            while(n % prime[i] == 0)  count ++,n /= prime[i];
            cnt = pow_mod(prime[i],count - 1);
            cnt *= solve2(prime[i]);
            ans = (ans / gcd(ans , cnt)) * cnt;
        }
    }
    if(n > 1){
        cnt = 1;
        cnt *= solve2(n);
        ans = ans / gcd(ans,cnt) * cnt;
    }
    return ans;
}
char Ss[30000007];
struct bign {
    int z[30000007],l;
    void init() {
 		memset(z,0,sizeof(z));
        scanf("%s",Ss + 1);
        l = strlen(Ss + 1);
        for(int i = 1;i <= l;i ++)
            z[i] = Ss[l - i + 1] - '0';
    }
    LL operator % (const long long & a) const {
        LL b = 0;
        for (int i = l;i >= 1;i --)
            b = (b * 10 + z[i]) % a;
        return b;
    }
}z;
LL m1;
struct Matrix {
    LL a[3][3];
    Matrix () { memset(a,0,sizeof a); }
    Matrix  operator * (const Matrix & p) const {
        Matrix ret;
        for(int i = 0;i <= 1;++ i)
            for(int j = 0;j <= 1;++ j)
                for(int k = 0;k <= 1;++ k)
                    ret.a[i][j] = (ret.a[i][j] + ( a[i][k] * p.a[k][j] ) % m1 ) % m1;
        return ret;
    }
};
LL solve(LL x) {
    Matrix p,q;
    p.a[0][0] = 1; p.a[0][1] = 1; p.a[1][0] = 1;
    q.a[0][1] = 1;
    for(;x;x >>= 1,p = p * p)
        if(x & 1) q = q * p;
    return q.a[0][0];
}
main() {
    LL t,n;
    init_prime();
    matrix_a.x1 = matrix_a.x2 = matrix_a.x3 = 1;
    matrix_a.x4 = 0;
    matrix_b.x1 = matrix_b.x4 = 1;
    matrix_b.x2 = matrix_b.x3 = 0;
    dp[2] = 3;dp[3] = 8;dp[5] = 20;
    z.init(); scanf("%lld",&n); m1 = n;
    n = get_M(n);
    //printf("%lld\n",n % m1);
    n = z % n;
    printf("%lld\n",solve(n));
    return 0;
}