POJ 2406 Power Strings

字符串Hash

对于一个字符串，枚举长度，在字符串中用$hash$进行比较

比如对于区间[l,r]，那么这一串字符的$hash$值为$hash[r]-base^{r-l+1}*hash[l-1]$

进行枚举即可

复杂度为$nsum_{i=1}^{n}frac{1}{i}$

由于$int frac{dx}{x}=ln x$

所以$sum_{i=1}^{n}frac{1}{i}leq ln n$

因此复杂度为$O(nln n)$

#include <bits/stdc++.h>
#define mod 1000000007
#define base 719
#define ll long long
using namespace std;
const ll MAXN=1001000;
char ch[MAXN];
ll ha[MAXN],z[MAXN];
int main()
{
    z[0]=1;
    for (ll i=1;i<=MAXN-10;i++)
      z[i]=(z[i-1]*base)%mod;
    while (1)
    {
        scanf("%s",ch+1);
        if (ch[1]=='.')
          break;
        ll len;
        len=strlen(ch+1);
        ha[0]=0;
        for (ll i=1;i<=len;i++)
          ha[i]=(ha[i-1]*base%mod+(ll)ch[i]%mod)%mod;
        ll aim;
        for (ll length=1;length<=len;length++)
        {
            aim=(ha[length]-(ha[0]*z[length]%mod)+mod)%mod;//目标
            if (len%length==0)
            {
                bool bl=1;
                for (ll r=length*2;r<=len;r+=length)
                {
                    ll h;
                    h=(ha[r]-(ha[r-length]*z[length]%mod)+mod)%mod;
                    if (aim!=h)
                    {
                        bl=0;
                        break;
                    }
                }
                if (bl)
                {
                    printf("%lld
",len/length);
                    break;
                }
            }
        }
    }
}