UVA - 10229 Modular Fibonacci 矩阵快速幂

                             Modular Fibonacci

The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...) are defined by the recurrence:
F0 = 0
F1 = 1
Fi = Fi−1 + Fi−2 for i > 1
Write a program which calculates Mn = Fn mod 2m for given pair of n and m. 0 ≤ n ≤ 2147483647
and 0 ≤ m < 20. Note that a mod b gives the remainder when a is divided by b.
Input
Input consists of several lines specifying a pair of n and m.
Output
Output should be corresponding Mn, one per line.
Sample Input
11 7
11 6
Sample Output
89
25

题解：

由于n<=2 147 483 647，直接for会超时。用矩阵快速幂就好了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;
ll MOD;
struct Matrix {
    ll mat[][];
}U,F;
Matrix multi (Matrix a, Matrix b) {
    Matrix ans;
    for(int i = ; i < ; i++) {
        for(int j = ; j < ; j++) {
            ans.mat[i][j] = ;
            for(int k = ; k < ; k++)
                ans.mat[i][j] += a.mat[i][k] * b.mat[k][j];
            ans.mat[i][j] %= MOD;
        }
    }
    return ans;
}
Matrix powss(ll n) {
    Matrix ans = U,p = F;
    while(n) {
        if(n&) ans = multi(ans,p);
        n>>=;
        p = multi(p,p);
    }
    return ans;
}
int main() {
    U = {,,,};
    F = {,,,};
    ll n,m;
    while(~scanf("%lld%lld",&n,&m)) {
        MOD = 1ll<<m;
        Matrix ans = powss(n);
        printf("%lld\n",ans.mat[][]);
    }
    return ;
}