UVa 10689 - Yet another Number Sequence

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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;
}

UVa 10689 - Yet another Number Sequence

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原文地址：http://blog.csdn.net/mobius_strip/article/details/45647965