HDU 4291 A Short problem

A Short problem

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4291
64-bit integer IO format: %I64d      Java class name: Main

 

　According to a research, VIM users tend to have shorter fingers, compared with Emacs users.
　　Hence they prefer problems short, too. Here is a short one:
　　Given $n (1 leq n leq 10^{18})$, You should solve for 
[g(g(g(n))) mod 10^9 + 7]
　　where
[g(n) = 3g(n - 1) + g(n - 2)]
[g(1) = 1]
[g(0) = 0]

 

Input

　　There are several test cases. For each test case there is an integer n in a single line.
　　Please process until EOF (End Of File).

 

Output

　　For each test case, please print a single line with a integer, the corresponding answer to this case.

 

Sample Input

0
1
2

Sample Output

0
1
42837

Source

2012 ACM/ICPC Asia Regional Chengdu Online

 

解题：矩阵快速幂，关键在于如何找到内部循环的的模对象

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 3;
LL mod = 1000000007;
const int n = 2;
struct Matrix {
    LL m[maxn][maxn];
    void init() {
        memset(m,0,sizeof m);
    }
    Matrix() {
        init();
    }
    Matrix operator*(const Matrix &rhs) const {
        Matrix ret;
        for(int k = 0; k < n; ++k) {
            for(int i = 0; i < n; ++i)
                for(int j = 0; j < n; ++j)
                    ret.m[i][j] = (ret.m[i][j] + m[i][k]*rhs.m[k][j]%mod)%mod;

        }
        return ret;
    }
    void set_a() {
        init();
        m[0][0] = 0;
        m[0][1] = 1;
    }
    void set_b() {
        init();
        m[0][0] = 0;
        m[0][1] = m[1][0] = 1;
        m[1][1] = 3;
    }
    void print() {
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < n; ++j)
                cout<<m[i][j]<<" ";
            cout<<endl;
        }
    }
};
Matrix a,b;
LL quickPow(LL index,LL md) {
    a.set_a();
    b.set_b();
    mod = md;
    while(index) {
        if(index&1) {
            a = a*b;
        }
        index >>= 1;
        b = b*b;
    }
    return a.m[0][0];
}
int main() {
    LL m;
    while(~scanf("%I64d",&m))
        printf("%I64d
",quickPow(quickPow(quickPow(m,183120),222222224),1000000007));
    return 0;
}