[codeforces-315D div2]模拟
题目：给两个字符串a、b，问从a中删去若干字符后最多可以得到多少个b串的重复串(bb...b的形式，b的长度不超过100)，其中a串是由一个长度不超过100的字符串s重复k次得到的
思路： 暴力匹配a和b，由于s,b的长度都不超过100，标记每次匹配后a串指针的位置对len(s)的模，那么最多有100种标记，每种标记最多导致a串指针移动100*100位，那么在a串的前1e6个字符，一定可以得到重复的标记，而重复的标记之间就是循环节，跳过中间的若干循环节，处理最后剩余的a串字符(一定小于1e6个)，继续暴力匹配下就行了。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 | /* ******************************************************************************** */#include <iostream> //#include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <vector> //#include <ctime> //#include <deque> //#include <queue> //#include <algorithm> //#include <map> //using namespace std; // //#define pb push_back //#define mp make_pair //#define X first //#define Y second //#define all(a) (a).begin(), (a).end() //#define foreach(a, i) for (typeof(a.begin()) i = a.begin(); i != a.end(); ++ i) //#define fill(a, x) memset(a, x, sizeof(a)) // //void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} //void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> //void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; //while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> //void print(const T t){cout<<t<<endl;}template<typename F,typename...R> //void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> //void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} // //typedef pair<int, int> pii; //typedef long long ll; //typedef unsigned long long ull; // //template<typename T>bool umax(T&a, const T&b){return b>a?false:(a=b,true);} //template<typename T>bool umin(T&a, const T&b){return b<a?false:(a=b,true);} //template<typename T> //void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} //template<typename T> //void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} // ///* -------------------------------------------------------------------------------- */ //const int maxn = 1234567;char s[maxn], a[123], c[123];int lena, lenc, b, d, ta;int buf[123], mark[123];void init() { char *ps = s; for (int i = 0; i < b; i ++) { for (int j = 0; j < lena; j ++) { *ps ++ = a[j]; if (ps - s == ta || ps - s > 1111111) return; } }}int work(int);void solve(int &cc, int pp) { if (pp == ta - 1) return ; int rest = ta - pp - 1, cnt = rest / lena; if (rest % lena) cnt ++; ta = cnt * lena; s[ta] = 0; cc += work(pp % lena + 1);}int work(int start) { memset(mark, 0, sizeof(mark)); memset(buf, 0, sizeof(buf)); int cc = 0, nowc = 0; for (int i = start; s[i]; i ++) { if (s[i] == c[nowc]) nowc ++; if (nowc == lenc) { cc ++; nowc = 0; if (mark[i % lena]) { cc += (ta - i - 1) / (i + 1 - mark[i % lena]) * (cc - buf[i % lena]); int pp = i + (ta - i - 1) / (i + 1 - mark[i % lena]) * (i + 1 - mark[i % lena]); solve(cc, pp); return cc; } else { mark[i % lena] = i + 1; buf[i % lena] = cc; } } } return cc;}int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE cin >> b >> d; scanf("%s%s", a, c); lena = strlen(a); lenc = strlen(c); ta = lena * b; init(); cout << work(0) / d << endl; return 0; //} // // // ///* ******************************************************************************** */ |
