HDU 4565 so easy(矩阵快速幂)

题目链接:?pid=4565

代码如下:

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <iostream>using namespace std;#define LL __int64struct matrix{LL f[2][2];};LL m;matrix mul(matrix a, matrix b){LL i, j, k;matrix c;memset(c.f, 0, sizeof(c.f));for (i = 0; i<2; i++)for (j = 0; j<2; j++)for (k = 0; k<2; k++)c.f[i][j] = (c.f[i][j] + a.f[i][k] * b.f[k][j]) % m;return c;}matrix pow_mod(matrix e, LL b){matrix s;s.f[0][0] = s.f[1][1] = 1;s.f[0][1] = s.f[1][0] = 0;while (b){if (b%2==1)s = mul(s, e);e = mul(e, e);b/=2;}return s;}int main(){LL a, b, n;while (scanf("%I64d%I64d%I64d%I64d", &a, &b, &n, &m)!=EOF) // while(cin>>a>>b>>n>>m){LL p, q, ans;matrix e;p = 2 * a;q = a*a – b;e.f[0][0] = p; e.f[0][1] = -q;e.f[1][0] = 1; e.f[1][1] = 0;e = pow_mod(e, n – 1);ans = ((p*e.f[0][0] + 2 * e.f[0][1]) % m + m) % m;printf("%d\n", ans);}return 0;}

,便觉不过如此。也许我们只是想让自己的心去旅行,

HDU 4565 so easy(矩阵快速幂)

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