UVa 10689 - Yet another Number Sequence

题目：给你Fib数列的前两项，求第n项的后m位的值。

分析：矩阵快速模幂。见本博客的：斐波那契数列

说明：╮(╯▽╰)╭。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>using namespace std;class matrix{private:int data[2][2];public:matrix(){};matrix(int a, int b, int c, int d){data[0][0] = a;data[0][1] = b;data[1][0] = c;data[1][1] = d;}matrix(int a, int b, int mod);friend matrix mul(matrix A, matrix B, int mod);friend matrix qpow(matrix mat, int n, int mod);int operator ()(int x, int y){return data[x-1][y-1];}};//矩阵乘法 matrix mul(matrix A, matrix B, int mod){matrix C;for (int i = 0 ; i < 2 ; ++ i)for (int j = 0 ; j < 2 ; ++ j) {C.data[i][j] = 0;for (int k = 0 ; k < 2 ; ++ k)C.data[i][j] = (C.data[i][j]+A.data[i][k]*B.data[k][j])%mod;}return C;}//矩阵快速幂 matrix qpow(matrix mat, int n, int mod){if (n == 1) return mat;matrix now = qpow(mat, n/2, mod);if (n%2 == 0) return mul(now, now, mod);return mul(mul(now, now, mod), mat, mod);}int mod[5] = {1, 10, 100, 1000, 10000};int main(){int t,a,b,n,m;while (cin >> t)while (t --) {cin >> a >> b >> n >> m;if (n == 0) cout << a%mod[m] << endl;else if (n == 1) cout << b%mod[m] << endl;else if (n == 2) cout << (a+b)%mod[m] << endl;else {matrix A(a, b, b, a+b);matrix B = qpow(matrix(0, 1, 1, 1), n-2, mod[m]);cout << mul(A, B, mod[m])(2,2) << endl;}}    return 0;}