MIPT-2016 Pre-Finals Workshop, Taiwan NTU Contest Fibonacci of Fibonacci

Fibonacci of Fibonacci

题目链接

题目大意：求第n项斐波那契数的斐波那契，即F_{Fn mod 20160519}

_{先找Fn mod 20160519的循环节,啊这个可以用板子求，然后用矩阵快速幂求一下就好了}

 

#include <cstdio>
#include <iostream>
#include <cstring>
typedef long long ll;
const int p = 20160519;
const ll mode = 26880696;//用板子求的，有求斐波那契循环节的题目
using namespace std;
ll mod;
struct mat {
    ll a[3][3];
    mat() {
        memset(a, 0, sizeof(a));
    }
    mat operator *(const mat &o) const {
        mat t;
        for(int i = 1; i <= 2; i++) {
            for(int j = 1; j <= 2; j++)
                for(int k = 1; k <= 2; k++) {
                    t.a[i][j] = (t.a[i][j] + a[i][k]*o.a[k][j])%mod;
                }
        }
        return t;
    }
} a, b;
ll slove(ll n) {
    a.a[1][1] = a.a[2][2] = 1, a.a[1][2] = a.a[2][1] = 0;
    b.a[1][1] = b.a[1][2] = b.a[2][1] = 1, b.a[2][2] = 0;
    if(n == 0) return 0;
    else {
        n--;
        while(n > 0) {
            if(n&1) {
                a = b*a;
                n--;
            }
            n >>= 1;
            b = b*b;
        }
       return a.a[1][1];
    }
}

int main() {
    int t;
    ll n;
    scanf("%d", &t);
    while(t--) {
        scanf("%lld", &n);
        mod = mode;
        ll ans = slove(n);
        mod = p;
        ans = slove(ans);
        printf("%lld
", ans);
    }
    return 0;
}