很显然是个进制转换的题，根据题意有a^2 = a + 1  -> a^n = a^(n-1) + a^(n-2)，这样就能消除两个连续1。

另，a^3 = a^2 + 1 = 2*a+2 = 2*(a+1)  =  2*a。这样就可以将悉数转化为01。

10^9大约是2^30,所以总长度不超高150，直接模拟就好了。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <ctime>
#include <iomanip>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-6)
#define LL long long
#define ULL unsigned long long
#define _LL __int64
#define INF 0x3f3f3f3f
#define Mod 1000000007
#define Seed 31

using namespace std;

ULL hash[100010];
ULL K;

char s[100010];

map<ULL,int> ma;

int main()
{
    int n,m,l,i,j;

    while(scanf("%d %d",&m,&l) != EOF)
    {
        scanf("%s",s);

        for(n = strlen(s), i = n-1 ,hash[n] = 0;i >= 0; --i)
            hash[i] = hash[i+1]*Seed + (s[i]-'a'+1);
        for(K = Seed,i = 2;i <= l; ++i)
            K *= Seed;
//        for(i = 0;i < n-l; ++i)
//            printf("i = %d hash = %I64u\n",i,hash[i]-hash[i+l]*K[l]);
        ULL tmp;
        int anw = 0;
        for(i = 0;i < l && m*l+i < n; ++i)
        {
            ma.clear();

            for(j = i;j < m*l+i;j += l)
                ma[hash[j] - hash[j+l]*K]++;

            if(ma.size() == m)
                anw++;

            for(j = m*l+i;j+l <= n; j += l)
            {
                ma[hash[j-m*l]-hash[j-m*l+l]*K]--;
                if(ma[hash[j-m*l]-hash[j-m*l+l]*K] == 0)
                    ma.erase(hash[j-m*l]-hash[j-m*l+l]*K);

                ma[hash[j] - hash[j+l]*K]++;

                if(ma.size() == m)
                    anw++;
            }
        }
        printf("%d\n",anw);
    }

    return 0;
}

HDU 4814 Golden Radio Base