poj 3070 Fibonacci 【矩阵快速幂 求第N个斐波那契数%1000】
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11123 | Accepted: 7913 |
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:
.
题意：给你一个求解第N个斐波那契数的公式。 让你求出Fn % 10000。
构造单位矩阵，然后就是矩阵快速幂了。
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 100
#define LL long long
#define MOD 10000
using namespace std;
struct Matrix
{
LL a[MAXN][MAXN];
int r, c;//行数 列数
};
Matrix ori, res;//初始矩阵 和 结果矩阵
void init()//初始化矩阵
{
memset(res.a, 0, sizeof(res.a));
res.r = 2; res.c = 2;
for(int i = 1; i <= 2; i++)//构造单位矩阵
res.a[i][i] = 1;
ori.r = 2; ori.c = 2;
ori.a[1][1] = ori.a[1][2] = ori.a[2][1] = 1;
ori.a[2][2] = 0;
}
Matrix multi(Matrix x, Matrix y)
{
Matrix z;
memset(z.a, 0, sizeof(z.a));
z.r = x.r, z.c = y.c;//新矩阵行数等于x矩阵的行数 列数等于y矩阵的列数
for(int i = 1; i <= x.r; i++)//x矩阵的行数
{
for(int k = 1; k <= x.c; k++)//矩阵x的列数等于矩阵y的行数 即x.c = y.r
{
if(x.a[i][k] == 0) continue;//优化
for(int j = 1; j<= y.c; j++)//y矩阵的列数
z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD;
}
}
return z;
}
void Matrix_mod(int n)
{
while(n)//N次幂
{
if(n & 1)
res = multi(ori, res);
ori = multi(ori, ori);
n >>= 1;
}
printf("%lld\n", res.a[1][2] % MOD);
}
int main()
{
int N;
while(scanf("%d", &N), N!=-1)
{
init();//初始化单位矩阵
Matrix_mod(N);//矩阵快速幂
}
return 0;
}