POJ 3070(矩阵快速幂）
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12195 | Accepted: 8656 |
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0 9 999999999 1000000000 -1
Sample Output
0 34 626 6875
Hint
As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by
.
Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:
.
Source
题意：求第n项的斐波那契数对10000求余
题解：快速矩阵幂，首先我们必须要构造出满足符合状态转移的基础矩阵，然而我并不会构造。。。。。。。。（感觉有点像DP推出状态转移方程）
接下来就矩阵乘法，但是明显n太大了，我们需要更高效的算法。这里可以想到使用快速幂加速乘法，我们可以使用倍增的思想去思考快速幂，这样可以在logn的复杂度内完成n次幂的乘法，（这里我终于是理解了快速幂的思想，可能是线段树做多了？）好啦这一题就先这样
这里献上我总结的模板：
#define MAXN 2
#define mod int(1e4)
struct Matrix
{
int mat[MAXN][MAXN];
Matrix() {}
Matrix operator*(Matrix const &b)const
{
Matrix res;
memset(res.mat, 0, sizeof(res.mat));
for (int i = 0 ;i < MAXN; i++)
for (int j = 0; j < MAXN; j++)
for (int k = 0; k < MAXN; k++)
res.mat[i][j] = (res.mat[i][j]+this->mat[i][k] * b.mat[k][j])%mod;
return res;
}
};
Matrix pow_mod(Matrix base, int n)
{
Matrix res;
memset(res.mat, 0, sizeof(res.mat));
for (int i = 0; i < MAXN; i++)
res.mat[i][i] = 1;
while (n > 0)
{
if (n & 1) res = res*base;
base = base*base;
n >>= 1;
}
return res;
}
AC代码：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<bitset>
#include<utility>
#include<functional>
#include<iomanip>
#include<sstream>
#include<ctime>
using namespace std;
#define MAXN 2
#define mod int(1e4)
struct Matrix
{
int mat[MAXN][MAXN];
Matrix() {}
Matrix operator*(Matrix const &b)const
{
Matrix res;
memset(res.mat, 0, sizeof(res.mat));
for (int i = 0 ;i < MAXN; i++)
for (int j = 0; j < MAXN; j++)
for (int k = 0; k < MAXN; k++)
res.mat[i][j] = (res.mat[i][j]+this->mat[i][k] * b.mat[k][j])%mod;
return res;
}
};
Matrix pow_mod(Matrix base, int n)
{
Matrix res;
memset(res.mat, 0, sizeof(res.mat));
for (int i = 0; i < MAXN; i++)
res.mat[i][i] = 1;
while (n > 0)
{
if (n & 1) res = res*base;
base = base*base;
n >>= 1;
}
return res;
}
int main()
{
#ifdef CDZSC
freopen("i.txt", "r", stdin);
//freopen("o.txt","w",stdout);
int _time_jc = clock();
#endif
Matrix base;
for (int i = 0; i < MAXN; i++)
for (int j = 0; j < MAXN; j++)
base.mat[i][j] = 1;
base.mat[1][1] = 0;
int n;
while (~scanf("%d", &n)&&n!=-1)
{
Matrix ans = pow_mod(base, n);
printf("%d\n", ans.mat[0][1]);
}
return 0;
}