字符串hash

简单字符串hash，通过对字符串每个位置进行加权运算，使得后面每个字符的hash值都受前面字符的影响，从而将多个字节的字符串转化为整数，但有时会发生冲突。

关键部分：seed权值，通常是质数，如果太小发生冲突的概率会大大增加。

  

具体代码：

typedef unsigned long long ull;
const int maxn=40005;
ull seed=31,ha[maxn],wei[amxn];

wei[0]=1;
for(int i=1;i<maxn;i++) wei[i]=wei[i-1]*seed;//对应长度字符串的权

ha[len]=0;
for(int i=len-1;i>=0;i--)
    ha[i]=ha[i+1]*seed+str[i]-'a';
//求长度为l的字符串哈希值
tmp=ha[i]-ha[i+l]*wei[l];

//对字符串hash可以从左到右，也可以从右到左
            

　　HDU4821MAP+HASH

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
#define N 100010

unsigned long long base=31,nbase[N],hash[N];
int n,len,ans,slen;
char str[N];
map<unsigned long long,int> mp;

int main()
{
    int i,j;
    unsigned long long tmp;
    nbase[0]=1;
    for(i=1;i<N;i++) nbase[i]=nbase[i-1]*base;
    while(~scanf("%d%d",&n,&len))
    {
        scanf("%s",str);
        slen=strlen(str);
        hash[slen]=0;
        for(i=slen-1;i>=0;i--)
        {
            hash[i]=hash[i+1]*base+str[i]-'a'+1;
        }
        ans=0;
        for(i=0;i<len&&i+n*len<=slen;i++)
        {
            mp.clear();
            for(j=i;j<i+n*len;j+=len)
            {
                tmp=hash[j]-hash[j+len]*nbase[len];
                mp[tmp]++;
            }
            if(mp.size()==n) ans++;
            for(j=i+n*len;j+len<=slen;j+=len)
            {
                tmp=hash[j-n*len]-hash[j-(n-1)*len]*nbase[len];
                mp[tmp]--;
                if(mp[tmp]==0) mp.erase(tmp);
                tmp=hash[j]-hash[j+len]*nbase[len];
                mp[tmp]++;
                if(mp.size()==n) ans++;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}

HDU4080二分+hash

#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
typedef unsigned long long uLL ;
const int maxn = 40050  ;
map<uLL,int>mp;
int pos ;
const uLL p = 9999997 ;
uLL powp[maxn] ;
uLL ha[maxn] ;
char s[maxn] ;

bool check(int l , int n , int m){
    bool ok = false ;
    mp.clear();
    for(int i=1; i+l-1<=n; i++) {
        uLL tmp = ha[i+l-1] - ha[i-1] * powp[l];
        if(++mp[tmp]>=m){pos=i;ok=true;}
    }
    return ok ;
}

int main()
{
    powp[0] = 1 ;
    for(int i=1; i<maxn; i++) powp[i] = powp[i-1] * p ;
    int m  ;
    while(scanf("%d" ,&m)==1 && m){
        scanf("%s" ,s+1) ;
        int n = strlen(s+1) ;
        ha[0] = ha[n+1] = 0 ;
        for(int i=1; i<=n;i++) ha[i] = ha[i-1]*p + s[i] ; //from end to begin ha  that's why s+1
        int L = 0 , R = n ;
        while(L<R) {
            int mid = L + (R-L+1)/2 ;
            if(check(mid , n ,m)) L = mid ;
            else R = mid - 1 ;
        }
        if(!L) puts("none") ;
        else printf("%d %d
" , L , pos-1) ;
    }
    return 0;
}