UVA10229Modular Fibonacci(矩阵快速幂)

题目链接

题目大意：给你i和m，求Mi， Mi = （F(i - 1) + F(i - 2)） % 2^m;

解题思路：因为Mi = (F（i - 1) % 2^m + F(i - 2)% 2^m) % 2^m = (M(i - 1) + M(i - 2)) % 2^m.类似于求fibonacci数加上取模，只是n很大，所以要用矩阵快速幂快速求解。参考

代码：

#include <cstdio>
#include <cstring>

typedef long long ll;
const int maxn = 2;
int N, M;

struct Mat {
    ll s[maxn][maxn];
    void init () {
        s[0][0] = s[0][1] = s[1][0] = 1;
        s[1][1] = 0;
    }

    Mat operator ^ (const Mat a) const{

        Mat ans;
        memset (ans.s, 0, sizeof (ans.s));

        for (int i = 0; i < maxn; i++)
            for (int j = 0; j < maxn; j++)
                for (int k = 0; k < maxn; k++)
                    ans.s[i][j] = (ans.s[i][j] + (s[i][k] * a.s[k][j]) % (1<<M)) % (1<<M);
        return ans;
    }
};

Mat FastMod (Mat a, int n) {

    if (n <= 1)
        return a;
    Mat tmp = FastMod (a, n / 2);
    tmp = tmp ^ tmp;
    if (n % 2 == 1)
        tmp = tmp ^ a;

    return tmp;
}

int main () {

    Mat a;
    a.init();
    while (scanf ("%d%d", &N, &M) != EOF) {

        if (!N) {
            printf ("0\n");
            continue;
        }
        Mat ans = FastMod (a, N);
        printf ("%lld\n", ans.s[1][0]);
    }
    return 0;
}