【POJ2406】Power Strings(hash/kmp)

题目地址:http://poj.org/problem?id=2406

题目:


求字符串是由多少个重复的子串组成的

 

解题思路:


kmp解法见:https://blog.csdn.net/Cassie_zkq/article/details/81665153

求出字符串的hash数组,从小到大枚举长度,和原字符串匹配,符合条件就输出。

 

ac代码:


#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef unsigned long long ll;
const ll maxn = 1000005;
ll p = 233333,len = 0;
char s[maxn];
ll has[maxn], power[maxn];
void init()//预处理p^n
{
    power[0] = 1;
    for(int  i = 1; i < maxn; i++)
        power[i] = power[i - 1] * p;
}
bool judge(int st)
{
    for(int i = st * 2; i <= len; i += st)
    {
        ll tmp = has[i] - has[i - st] * power[st];
        if(tmp != has[st]) return false;
    }
    return true;
}
int main ()
{
    //freopen("/Users/zhangkanqi/Desktop/11.txt","r",stdin);
    init();
    while(scanf("%s",s+1))
    {
        if(s[1] == '.') break;
        len = strlen(s + 1);
        has[0] = 0;
        for(int i = 1; i <= len; i++)
            has[i] = has[i - 1] * p + (ll)(s[i] - 'a' + 1);
        for(int i = 1; i <= len ; i++)
        {

            if(len % i == 0 && judge(i))
            {
                printf("%llu\n", len / i);
                break;
            }
        }
    }
    return 0;
}

 

你可能感兴趣的:(字符串hash)