Symmetric length of prefix and suffix (double hash)

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Given a number of strings (the total length of these strings <=4*10^5), find the length of all substrings that are both prefixes and suffixes in each string.

For example: aabbcababababcabab, which is both a prefix and a suffix: ab, abab, ababcabab, ababcababababcabab.

Input format

Enter several lines, one string per line.

Output format

For each string, output a line containing several incrementing integers, indicating the length of all substrings that are both prefix and suffix.

Sample input

ababcababababcabab
aaaaa

Sample output

2 4 9 18
1 2 3 4 5

Algorithm: Double Hash

Code

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N	400100
#define B   131
using namespace std;
typedef long long ll;
ll sum1[N],sum2[N];
ll mod1=1e9+7,mod2=1e9+9;
ll power1[N],power2[N];
char s[N];
int len;
/*双哈希，数组sum1，sum2分别存放这哈希值
同时右分别对mod1,mod2取模，
使用这两个孪生质数再用双哈希
就几乎不可能发生冲突

数组power用来存幂
B用来表示幂的底数 
*/
void init()
{
	power1[0]=1;power2[0]=1; 
	for(int i=1;i<=N;i++)
	{
		power1[i]=(power1[i-1]*B)%mod1;
		power2[i]=(power2[i-1]*B)%mod2;	
	}
} 
int judge(int x)
//x表示前缀和后缀所要比的长度 
{
	ll lead1=sum1[x];
	ll train1=((sum1[len]-sum1[len-x]*power1[x]%mod1)%mod1+mod1)%mod1;
	ll lead2=sum2[x];                              //%mod1不可少 
	ll train2=((sum2[len]-sum2[len-x]*power2[x])%mod2+mod2)%mod2;
	if(lead1==train1&&lead2==train2)
		return 1;
	return 0;
}
int main()
{
	init();//预处理 
	while(~scanf("%s",s+1))
	{
		len=strlen(s+1);
		sum1[0]=0;sum2[0]=0;
		for(int i=1;i<=len;i++)
		{
			sum1[i]=(sum1[i-1]*B+s[i])%mod1;
			sum2[i]=(sum2[i-1]*B+s[i])%mod2;
		}
	
		for(int i=1;i<=len;i++)
		{
			if(judge(i)==1)
				printf("%d ",i);
		}
		printf("\n");
	} 

	return 0;
}