一块花布条,里面有些图案,另有一块直接可用的小饰条,里面也有一些图案。对于给定的花布条和小饰条,计算一下能从花布条中尽可能剪出几块小饰条来呢?
Input输入中含有一些数据,分别是成对出现的花布条和小饰条,其布条都是用可见ASCII字符表示的,可见的ASCII字符有多少个,布条的花纹也有多少种花样。花纹条和小饰条不会超过1000个字符长。如果遇见#字符,则不再进行工作。
Output输出能从花纹布中剪出的最多小饰条个数,如果一块都没有,那就老老实实输出0,每个结果之间应换行。
Sample Input
abcde a3 aaaaaa aa #
Sample Output
0 3
#include <bits/stdc++.h> #include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> #include<queue> #include<map> #include<set> #include<vector> #include<iomanip> using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const int mod = 998244353; const double eps = 1e-8; const int mx = 1005; //check the limits, dummy typedef pair<int, int> pa; const double PI = acos(-1); ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } ll lcm(ll a, ll b) { return a * b / gcd(a, b); } bool isprime(int n) { if (n <= 1)return 0; for (int i = 2; i * i <= n; i++)if (n % i == 0)return 0; return 1; } #define swa(a,b) a^=b^=a^=b #define re(i,a,b) for(int i=(a),_=(b);i<_;i++) #define rb(i,a,b) for(ll i=(a),_=(b);i>=_;i--) #define clr(a,b) memset(a, b, sizeof(a)) #define lowbit(x) ((x)&(x-1)) #define mkp make_pair //inline ll qpow(ll a, ll b) { return b ? ((b & 1) ? a * qpow(a * a % mod, b >> 1) % mod : qpow(a * a % mod, b >> 1)) % mod : 1; } //inline ll qpow(ll a, ll b, ll c) { return b ? ((b & 1) ? a * qpow(a * a % c, b >> 1) % c : qpow(a * a % c, b >> 1)) % c : 1; } void ca(int kase, int ans) { cout << "Case #" << kase << ": " << ans << endl; } void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); } int n, m, t, k; char str[mx], pattern[mx]; int Next[mx]; int cnt; void getFail(char* p, int plen) { Next[0] = 0, Next[1] = 0; re(i, 1, plen) { int j = Next[i]; while (j && p[i] != p[j])j = Next[j]; Next[i + 1] = (p[i] == p[j]) ? j + 1 : 0; } } void kmp(char* s, char* p) { int last = -1; int slen = strlen(s), plen = strlen(p); getFail(p,plen); int j = 0; re(i, 0, slen) { while (j && s[i] != p[j])j = Next[j]; if (s[i] == p[j])j++; if (j == plen) { if (i - last >= plen)cnt++, last = i; } } } int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); while (cin >> str) { if (str[0] == '#')break; cin >> pattern; cnt = 0; kmp(str, pattern); cout << cnt << endl; } return 0; }