Poj 1200 --- a good string hash Construction Method

Question: http://poj.org/problem? Id = 1200

For a string, given n and NC, the string contains a maximum of NC characters. The maximum number of different substrings with N length can be:

Like the one in the last field competition, it is also difficult to judge and rehandle the problem and it will time out.

Method: map a substring with a length of N to an NC number, and open a large array to check whether the substring is repeated.

#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <cmath>#include <map>#include <map>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdin)const int MAXN = 16000000+100;bool Hash[MAXN];char str[MAXN];int m[1001];int n,nc;int ah,base;///int main(){    //IN("poj1200.txt");    int ans,len,seed;    while(~scanf("%d%d",&n,&nc))    {        CL(Hash,0);CL(m,0);        ah=0;ans=0;seed=0;        scanf("%s",str);        len=strlen(str);        if(len<n){puts("0");continue;}        rep(i,0,len)        {            if(!m[str[i]])m[str[i]]=++seed;            if(seed == nc)break;        }        ah=m[str[0]],base=nc;        for(int i=1;i<n;i++)        {            ah=ah*nc+m[str[i]];            base*=nc;        }        Hash[ah]=1;ans++;        rep(i,n,len)        {            ah=ah*nc-m[str[i-n]]*base+m[str[i]];            if(!Hash[ah])            {                Hash[ah]=1;                ans++;            }        }        printf("%d\n",ans);    }    return 0;}