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  • [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<intint> 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;                                                                       //
    }                                                                                   //
                                                                                        //
                                                                                        //
                                                                                        //
    /* ******************************************************************************** */
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  • 原文地址:https://www.cnblogs.com/jklongint/p/4687970.html
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