pek （北大oj）3070

思路：矩阵快速幂， 二分加速

#include<cstdio>
#include<cstring>
#define ll long long
#define mod 10000

struct jz
{
    ll num[2][2];
    jz(){ memset(num, 0, sizeof(num)); }
    jz operator*(const jz&p)const
    {
        jz ans;
        for (int k = 0; k < 2;++k)
        for (int i = 0; i < 2;++i)
        for (int j = 0; j < 2; ++j)
            ans.num[i][j] = (ans.num[i][j] + num[i][k] * p.num[k][j] % mod) % mod;
        return ans;
    }
};
jz POW(jz x, ll n)
{
    jz ans;
    for (int i = 0; i < 2; ++i)ans.num[i][i] = 1;
    for (; n; n>>=1, x=x*x)
    if (n & 1){ ans = ans*x; }
    return ans;
}
int main()
{
    ll n;
    while (scanf("%lld", &n) && (n != -1))
    {
        if (n <= 1)printf("%lld\n", n);
        else
        {
            jz pp;
            pp.num[0][0] = 1; pp.num[0][1] = 1;
            pp.num[1][0] = 1;
            pp = POW(pp, n - 2);
            printf("%lld\n", (pp.num[0][0]+pp.num[0][1])%mod);
        }
    }
}