哈希 poj2752 Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17568		Accepted: 9009

Description

The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm:

Step1. Connect the father's name and the mother's name, to a new string S.
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).

Example: Father='ala', Mother='la', we have S = 'ala'+'la' = 'alala'. Potential prefix-suffix strings of S are {'a', 'ala', 'alala'}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:)

Input

The input contains a number of test cases. Each test case occupies a single line that contains the string S described above.

Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.

Output

For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby's name.

Sample Input

ababcababababcabab
aaaaa

Sample Output

2 4 9 18
1 2 3 4 5

Source

POJ Monthly--2006.01.22,Zeyuan Zhu

第一道哈希模板题，WA了好多发，写个题解吧。

首先做法很简单，哈希预处理，枚举长度即可。

下面是需要注意的点：

1.哈希的mod的处理，选一个比较好的数，而且注意运算过程中取模运算，同时注意负数取模

2.expp【】数组要开成字符串规模大小！

#include
#include
#include
#include
#include
#include
#include
using namespace std;
const long long MOD=1000000000+27;
char s[400000+10];
long long a[400000+10],n,expp[400000+10];//expp[i] means 31^i
void init(void)
{
    expp[0]=1;
    for(int i=1;i<400000+10;i++)
        expp[i]=(expp[i-1]*31)%MOD;
}
int main(void)
{
    init();
    while(scanf("%s",s+1)!=EOF)
    {
        memset(a,0,sizeof(a));
        a[0]=0;
        n=strlen(s+1);
        a[1]=(s[1]-'a'+1);
        for(int i=2;i<=n;i++)
        {
            a[i]=(a[i-1]+expp[i-1]*(s[i]-'a'+1))%MOD;
        }
        for(int i=1;i<=n;i++)
        {
            long long tempa=(a[i])%MOD,tempb=(a[n]-a[n-i]+MOD)%MOD;
            if((tempa*expp[n-i])%MOD==tempb)
                printf("%d ",i);
        }
        printf("\n");
    }
    return 0;
}

3.每一位的权值不要是0，否则0000 与0 无法分辨。