CodeForces 450B Jzzhu and Sequences
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矩阵快速幂。
首先得到公式
然后构造矩阵,用矩阵加速
取模函数需要自己写一下,是数论中的取模。
#include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> using namespace std; long long MOD = 1e9 + 7; long long x, y; int n; long long mod(long long a, long long b) { if (a >= 0) return a%b; if (abs(a) % b == 0) return 0; return (a + b*(abs(a) / b + 1)); } struct Matrix { long long A[5][5]; int R, C; Matrix operator*(Matrix b); }; Matrix X, Y, Z; Matrix Matrix::operator*(Matrix b) { Matrix c; memset(c.A, 0, sizeof(c.A)); int i, j, k; for (i = 1; i <= R; i++) for (j = 1; j <= C; j++) for (k = 1; k <= C; k++) c.A[i][j] = mod((c.A[i][j] + mod(A[i][k] * b.A[k][j], MOD)), MOD); c.R=R; c.C=b.C; return c; } void read() { scanf("%lld%lld%d", &x, &y, &n); } void init() { n = n - 1; Z.A[1][1] = x, Z.A[1][2] = y; Z.R = 1; Z.C = 2; Y.A[1][1] = 1, Y.A[1][2] = 0, Y.A[2][1] = 0, Y.A[2][2] = 1; Y.R = 2; Y.C = 2; X.A[1][1] = 0, X.A[1][2] = -1, X.A[2][1] = 1, X.A[2][2] = 1; X.R = 2; X.C = 2; } void work() { while (n) { if (n % 2 == 1) Y = Y*X; n = n >> 1; X = X*X; } Z = Z*Y; printf("%lld\n", mod(Z.A[1][1], MOD)); } int main() { read(); init(); work(); return 0; }
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