UVA10229Modular Fibonacci(矩阵快速幂)

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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;}

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