CF119D（字符串-哈希求解（KMP求了半天，结果哈希更简单！））

做提前一定要先比较KMP和哈希哪个更简单！！

Let s be a string whose length equals n. Its characters are numbered from 0 to n - 1, i and j are integers, 0 ≤ i < j < n. Let's define function f as follows:

f(s, i, j) = s[i + 1... j - 1] + r(s[j... n - 1]) + r(s[0... i]).

Here s[p... q] is a substring of string s, that starts in position p and ends in position q (inclusive); "+" is the string concatenation operator; r(x) is a string resulting from writing the characters of the x string in the reverse order. If j = i + 1, then the substring s[i + 1... j - 1] is considered empty.

You are given two strings a and b. Find such values of i and j, that f(a, i, j) = b. Number i should be maximally possible. If for this i there exists several valid values of j, choose the minimal j.

Input

The first two input lines are non-empty strings a and b correspondingly. Each string's length does not exceed 106characters. The strings can contain any characters with ASCII codes from 32 to 126 inclusive.

Output

Print two integers i, j — the answer to the problem. If no solution exists, print "-1 -1" (without the quotes).

Sample test(s)

input

Die Polizei untersucht eine Straftat im IT-Bereich.
untersucht eine Straftat.hciereB-TI mi  ieziloP eiD

output

11 36

input

cbaaaa
aaaabc

output

4 5

input

123342
3324212

output

-1 -1

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<vector>
#define tree int o,int l,int r
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define lo o<<1
#define ro o<<1|1
#define ULL unsigned long long
#define LL long long
#define UI unsigned int
#define inf 0x7fffffff
#define eps 1e-7
#define N 1000009
#define f 97531997
using namespace std;
int T,n,m,k,lena,lenb,x,y,ok;
char a[N],b[N];
ULL ha[N],fa[N],hb[N],ep[N];
int ll[N],rr[N];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("ex.in","r",stdin);
#endif
    int ncase=0;
    while (gets(a+1))
    {
        gets(b+1);
        lena=strlen(a+1);
        lenb=strlen(b+1);
        if(lena!=lenb)
        {
            puts("-1 -1");continue;
        }
        n=lena;
        ha[0]=fa[0]=hb[0]=0;
        ep[0]=1;
        for(int i=1;i<=n;i++)
        {
            ep[i]=ep[i-1]*f;
            ha[i]=ha[i-1]*f+a[i];
            hb[i]=hb[i-1]*f+b[i];
            fa[i]=fa[i-1]*f+a[n-i+1];
        }
        for(int i=1;i<=n;i++)
        {
            int l=0,r=n+1-i;
            while(l<r)//二分求长度！
            {
                int mid=(l+r+1)>>1;
                if(fa[mid]==hb[i+mid-1]-hb[i-1]*ep[mid])
                l=mid;
                else
                r=mid-1;
            }
            rr[i]=l;
        }
        for(int i=1,j=1;i<n;i++)
        {
            for(;j<i+rr[i];j++)
            ll[j]=i;
        }
        for(int i=1;i<n;i++)
        {
            if(a[i]!=b[n+1-i])break;
            if(ll[n-i]>n-i||ll[n-i]==0)continue;//WA,ll[n-i]==0

            int len=n-i-ll[n-i]+1;
            if(hb[ll[n-i]-1]==ha[n-len]-ha[i]*ep[n-i-len])
            {
                x=i,y=n-len+1;
            }
        }
        printf("%d %d\n",x-1,y-1);
    }
    return 0;
}