Sol：就是求第N项的斐波那契数。矩阵乘法+快速幂

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>

using namespace std;

#define LL long long

struct Mat{
    LL f[2][2];
};

LL MOD = 10000;

Mat mul(Mat a,Mat b)
{
    LL i,j,k;
    Mat c;
    memset(c.f,0,sizeof(c.f));
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
            for(k=0;k<2;k++)
            c.f[i][j]=(c.f[i][j]+a.f[i][k]*b.f[k][j])%MOD;//可以改为不%MOD 
    return c;
}

Mat pow_mod(Mat e,LL b)
{
    Mat s;
    s.f[0][0]=s.f[1][1]=1;
    s.f[0][1]=s.f[1][0]=0;
    while(b)
    {
        if(b&1)
            s=mul(s,e);
        e=mul(e,e);
        b=b>>1;
    }
    return s;
}
int main()
{
    LL a,b,n,m;
    //while(~scanf("%lld%lld%lld%lld",&a,&b,&n,&m))
   // while(~scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m))
    while(~scanf("%I64d",&n),n+1)
	{
		if(n==0)
		{
			printf("0\n");
			continue;
		}
		n--;
        LL ans;
        Mat e;
        e.f[0][0]=1;e.f[0][1]=1;
        e.f[1][0]=1;e.f[1][1]=0;
        e=pow_mod(e,n-1);
        ans=((e.f[0][0]+e.f[1][0])%MOD+MOD)%MOD;   //可能负数结果 
        //printf("%lld\n",ans);
        printf("%I64d\n",ans);
    }
    return 0;
}