uva 10229 - Modular Fibonacci(矩阵快速幂）

题目大意：给出n和m，求出f(n) %m, f(x)为斐波那契数列。

解题思路：因为n的范围在0~214783647，所以计算量比较大，所以用矩阵快速幂。

{(1, 1), (1, 0)} ^ n *（f[1], f[0]) = (f[n], f[n - 1]).

#include <stdio.h>  
#include <math.h>  
long long n, b, k;  
  
struct state {  
    long long s[2][2];  
    state(long long a = 0, long long b = 0, long long  c = 0, long long d = 0) {  
        s[0][0] = a, s[0][1] = b, s[1][0] = c, s[1][1] = d;  
    }  
}tmp(1, 0, 0, 1), c(1, 1, 1, 0);  
  
state count(const state& p, const state& q) {  
    state f;  
    for (int i = 0; i < 2; i++)  
        for (int j = 0; j < 2; j++)  
            f.s[i][j] = (p.s[i][0] * q.s[0][j] + p.s[i][1] * q.s[1][j]) % b;  
    return f;  
}  
  
state solve(long long k) {  
    if (k == 0) return tmp;  
    else if (k == 1) return c;  
  
    state a = solve(k / 2);  
  
    a = count(a, a);  
    if (k % 2) a = count(a, c);  
    return a;  
}  
  
int main () {  
    int m;  
    while (scanf("%lld%d", &n, &m) == 2) {  
        b = pow(2, m);  
        if (n) {  
            state ans = solve(n - 1);  
            k = ans.s[0][0];  
        } else  
            k = 0;  
        printf("%lld\n", k);  
    }  
    return 0;  
}

补充：软件开发 , C++ ,