[hdu2087]kmp水题

题意：求模板串在文本串中出现的次数（位置无交叉）。只需在找到的时候把模板串指针归0即可。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d
", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

typedef double db;
typedef long long LL;
typedef pair<int, int> pii;
typedef multiset<int> msi;
typedef set<int> si;
typedef vector<int> vi;
typedef map<int, int> mii;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 2e5 + 7;
const int md = 10007;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

struct KMP {
    int next[1000010];
    void GetNext(char s[]) {
        mem0(next);
        next[0] = next[1] = 0;
        for(int i = 1; s[i]; i++) {
            int j = next[i];
            while(j && s[i] != s[j]) j = next[j];
            next[i + 1] = s[j] == s[i]? j + 1 : 0;
        }
    }
    int find(char s[], char s2[]) {
        int j = 0, len = strlen(s2), ans = 0;
        for (int i = 0; s[i]; i++) {
            while (j && s[i] != s2[j]) j = next[j];
            if (s[i] == s2[j]) j++;
            if (j == len) {
                ans++;
                j = 0;
            }
        }
        return ans;
    }
};

KMP kmp;
char s[maxn], s2[maxn];

int main() {
    //freopen("in.txt", "r", stdin);
    while (scanf("%s", s)) {
        if (s[0] == '#') break;
        scanf("%s", s2);
        kmp.GetNext(s2);
        cout << kmp.find(s, s2) << endl;
    }
    return 0;
}