POJ

题目链接

http://poj.org/problem?id=3070

题意

求斐波那契数列的第N项%10000

思路
因为 N 是一个非常大的数 如果一步一步 递推 是会T的
如果打表 是会M的

所以采用矩阵快速幂的方法

单位矩阵 为

1 0
0 1

n == 0 的时候 就是 单位矩阵

n == 1 的时候 就是 单位矩阵 * 1 1
1 0
依次类推 a[0][1] 或者 a[1][0] 就是当前项

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a, b) memset(a, (b), sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = acos(-1.0);
const double E = exp(1.0);
const double eps = 1e-8;

const int INF = 0x3f3f3f3f;
const int maxn = 1e2 + 5;
const int MOD = 1e4;

struct Matrix
{
    int a[2][2];
    Matrix () {}
    Matrix operator * (Matrix const &b)const
    {
        Matrix res;
        CLR(res.a, 0);
        for (int i = 0; i < 2; i++)
            for (int j = 0; j < 2; j++)
                for (int k = 0; k < 2; k++)
                    res.a[i][j] = (res.a[i][j] + this->a[i][k] * b.a[k][j]) % MOD;
        return res;
    }
};

Matrix pow_mod(Matrix base, int n)
{
    Matrix ans;
    CLR(ans.a, 0);
    for (int i = 0; i < 2; i++)
        ans.a[i][i] = 1;
    while (n > 0)
    {
        if (n & 1)
            ans = ans * base;
        base = base * base;
        n >>= 1;
    }
    return ans;
}


int main()
{
    Matrix base;
    base.a[0][0] = base.a[0][1] = base.a[1][0] = 1;
    base.a[1][1] = 0;
    int n;
    while (scanf("%d", &n) && n != -1)
    {
        Matrix ans = pow_mod(base, n);
        cout << ans.a[0][1] << endl;
    }
}