POJ

题目已经告诉你斐波那契矩阵算法了...

所以需要用到矩阵快速幂，看起来名字很diao但其实和普通快速幂基本一毛一样...

#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <set>
#include <stack>
#include <math.h>
#include <string>
#include <algorithm>

#define SIGMA_SIZE 26
#define pii pair<int,int>
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) (x&-x)
#define fode(i, a, b) for(int i=a; i>=b; i--)
#define foe(i, a, b) for(int i=a; i<=b; i++)
#define fod(i, a, b) for(int i=a; i>b; i--)
#define fo(i, a, b) for(int i=a; i<b; i++)
//#pragma warning ( disable : 4996 )

using namespace std;
typedef long long LL;
inline LL LMax(LL a, LL b) { return a>b ? a : b; }
inline LL LMin(LL a, LL b) { return a>b ? b : a; }
inline LL lgcd(LL a, LL b) { return b == 0 ? a : lgcd(b, a%b); }
inline LL llcm(LL a, LL b) { return a / lgcd(a, b)*b; }  //a*b = gcd*lcm
inline int Max(int a, int b) { return a>b ? a : b; }
inline int Min(int a, int b) { return a>b ? b : a; }
inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b) { return a / gcd(a, b)*b; }  //a*b = gcd*lcm
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL lmod = 1e9+7;
const int mod = 10000;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int maxk = 1e5+5;
const int maxm = 510*510;
const int maxn = 2;

struct Matrix {
    int mm[maxn][maxn];
    Matrix() {}
    Matrix operator*(Matrix const &b) const
    {
        Matrix tmp; memset(tmp.mm, 0, sizeof(tmp.mm));
        fo(i, 0, maxn)
            fo(j, 0, maxn)
                fo(k, 0, maxn)
                    tmp.mm[i][j] = (tmp.mm[i][j] + this->mm[i][k]*b.mm[k][j])%mod;
        return tmp;
    }
};

Matrix qpow(Matrix base, int b)
{
    Matrix res; 
    memset(res.mm, 0, sizeof(res.mm));
    fo(i, 0, maxn) res.mm[i][i] = 1;
    while(b)
    {
        if (b&1) res = res*base;
        base = base*base;
        b >>= 1;
    }
    return res;
}


int main()
{

    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
    #endif
    
    int x;
    while(cin >> x)
    {
        if (x==-1) break;
        //if (x==0) cout << 0 << endl;

        Matrix tmp, ans; memset(tmp.mm, 0, sizeof(tmp.mm));
        tmp.mm[0][0] = tmp.mm[0][1] = tmp.mm[1][0] = 1;
        tmp.mm[1][1] = 0;

        ans = qpow(tmp, x);
        printf("%d
", ans.mm[0][1]);
    }

    return 0;
}