[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。

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/* ******************************************************************************** */ #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;                                                                       // }                                                                                   //                                                                                     //                                                                                     //                                                                                     // /* ******************************************************************************** */