noip模拟赛 残

分析：这道题有点丧病啊......斐波那契数列本来增长就快，n <= 10^100又套2层，看到题目就让人绝望.不过这种题目还是有套路的.首先求斐波那契数列肯定要用到矩阵快速幂，外层的f可以通过取模来变小，可是里面的f不能直接取模1e9+7.因为余数最多就1e9+7种，所以肯定有一个循环节，打表发现内层f的循环节是2000000016,x的循环节是(1e9+7)*3,在求得时候mod循环节长度就ok了.

关于斐波那契的一些套路要记住：用矩阵快速幂加速、有循环节......

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const long long mod = ;
const long long mod2 = mod *  + ;
const long long mod3 = mod2 * ;

typedef long long ll;

int T, len, shu[];
char s[];
ll t;

struct node
{
    ll a[][];
    node(){ memset(a, , sizeof(a)); }
}x, y;

ll zhuanhuan()
{
    ll res = ;
    for (int i = ; i <= len; i++)
        shu[i] = s[i] - '';
    for (int i = ; i <= len; i++)
        res = (res *  + shu[i]) % mod3;
    return res;
}

node mul1(node x, node y)
{
    node p;
    memset(p.a, , sizeof(p.a));
    for (int i = ; i <= ; i++)
        for (int j = ; j <= ; j++)
            for (int k = ; k <= ; k++)
                p.a[i][j] = (p.a[i][j] + x.a[i][k] * y.a[k][j] % mod2) % mod2;
    return p;
}

node mul2(node x, node y)
{
    node p;
    memset(p.a, , sizeof(p.a));
    for (int i = ; i <= ; i++)
        for (int j = ; j <= ; j++)
            for (int k = ; k <= ; k++)
                p.a[i][j] = (p.a[i][j] + x.a[i][k] * y.a[k][j] % mod) % mod;
    return p;
}

ll qpow1(ll b)
{
    x.a[][] = ;
    x.a[][] = ;
    y.a[][] = ;
    y.a[][] = y.a[][] = y.a[][] = ;
    while (b)
    {
        if (b & )
            x = mul1(x, y);
        y = mul1(y, y);
        b >>= ;
    }
    return x.a[][];
}

ll qpow2(ll b)
{
    x.a[][] = ;
    x.a[][] = ;
    y.a[][] = ;
    y.a[][] = y.a[][] = y.a[][] = ;
    while (b)
    {
        if (b & )
            x = mul2(x, y);
        y = mul2(y, y);
        b >>= ;
    }
    return x.a[][];
}

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%s", s + );
        len = strlen(s + );
        t = zhuanhuan();
        t = qpow1(t);
        printf("%lld\n", qpow2(t));
    }

    return ;
}