HDU-3336-Count the string(扩展KMP)

链接:

https://vjudge.net/problem/HDU-3336

题意:

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:

s: "abab"

The prefixes are: "a", "ab", "aba", "abab"

For each prefix, we can count the times it matches in s. So we can see that prefix "a" matches twice, "ab" matches twice too, "aba" matches once, and "abab" matches once. Now you are asked to calculate the sum of the match times for all the prefixes. For "abab", it is 2 + 2 + 1 + 1 = 6.

The answer may be very large, so output the answer mod 10007.

思路:

计算s前缀出现的次数, 考虑扩展KMP的Nex数组, 从i位置开始从s0开始匹配的长度.

即这些长度的前缀都出现过.累加和即可.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#include <iostream>
#include <sstream>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 2e5+10;
const int MOD = 1e4+7;

char s1[MAXN], s2[MAXN];
int Next[MAXN], Exten[MAXN];

void GetNext(char *s)
{
    int len = strlen(s);
    int a = 0, p = 0;
    Next[0] = len;
    for (int i = 1;i < len;i++)
    {
        if (i >= p || i+Next[i-a] >= p)
        {
            if (i >= p)
                p = i;
            while (p < len && s[p] == s[p-i])
                p++;
            Next[i] = p-i;
            a = i;
        }
        else
            Next[i] = Next[i-a];
    }
}

void ExKmp(char *s, char *t)
{
    int len = strlen(s);
    int a = 0, p = 0;
    GetNext(t);
    for (int i = 0;i < len;i++)
    {
        if (i >= p || i + Next[i-a] >= p)
        {
            if (i >= p)
                p = i;
            while (p < len && s[p] == t[p-i])
                p++;
            Exten[i] = p-i;
            a = i;
        }
        else
            Exten[i] = Next[i-a];
    }

}

int main()
{
    int t, n;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        scanf("%s", s1);
        GetNext(s1);
        int res = 0;
        for (int i = 0;i < strlen(s1);i++)
            res = (res+Next[i])%MOD;
        printf("%d\n", res);
    }

    return 0;
}