HDU 6194string string string

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HDU 6194

输入一个字符串，求出现K次得子串个数

我的做法是求出大于等于K次的子串个数-大于等于K+1次的子串个数

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <string>#include <stdlib.h>using namespace std;typedef long long LL;const int maxn = 1e5 + 10;const int maxx = 127000;struct Node {    int pre, nxt[26];    int step, number1, num;    void Clear() {        step = number1 = num = 0;        pre = -1;        memset(nxt, -1, sizeof(nxt));    }} st[maxn<<1];int t, root, last, cur, k;LL ans, tmp;char str[maxn];void init() {    cur = 0;    root = last = cur++;    st[root].Clear();}void extend(int w) {    int np = cur++, p = last;    last = np;    st[np].Clear();    st[np].step = st[p].step+1;    while(p != -1 && st[p].nxt[w] == -1)        st[p].nxt[w] = np, p = st[p].pre;    if(p == -1)        st[np].pre = root;    else {        int q = st[p].nxt[w];        if(st[q].step == st[p].step+1)            st[np].pre = q;        else {            int nq = cur++;            st[nq].Clear();            st[nq].number1 = st[q].number1;            st[nq].num = st[q].num;            memcpy(st[nq].nxt, st[q].nxt, sizeof(st[q].nxt));            st[nq].pre = st[q].pre;            st[nq].step = st[p].step+1;            st[np].pre = st[q].pre = nq;            while(p != -1 && st[p].nxt[w] == q)                st[p].nxt[w] = nq, p = st[p].pre;        }    }    int f1 = 0, f2 = 0, pp = np;    while(np!=-1 && (!f1 || !f2)){        if(f1 == 0 && st[np].number1 < k) {            st[np].number1++;            if(st[np].number1 == k){                ans+=(LL)st[np].step-st[st[np].pre].step;                f1 = 1;            }        }        if(f2 == 0 && st[np].num < k+1) {            st[np].num++;            if(st[np].num == k+1){                tmp+=(LL)st[np].step-st[st[np].pre].step;                f2 = 1;            }        }        np = st[np].pre;    }}void solve() {    scanf("%d", &t);    while(t--) {        scanf("%d%s", &k, str);        init();        ans = tmp = 0;        int len = strlen(str);        for(int i = 0; i < len; i++)            extend(str[i]-'a');        printf("%lld\n", ans-tmp);    }}int main() {    solve();}

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