zoukankan      html  css  js  c++  java
  • hdu 4821

    String

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1410    Accepted Submission(s): 414


    Problem Description
    Given a string S and two integers L and M, we consider a substring of S as “recoverable” if and only if
      (i) It is of length M*L;
      (ii) It can be constructed by concatenating M “diversified” substrings of S, where each of these substrings has length L; two strings are considered as “diversified” if they don’t have the same character for every position.

    Two substrings of S are considered as “different” if they are cut from different part of S. For example, string "aa" has 3 different substrings "aa", "a" and "a".

    Your task is to calculate the number of different “recoverable” substrings of S.
     
    Input
    The input contains multiple test cases, proceeding to the End of File.

    The first line of each test case has two space-separated integers M and L.

    The second ine of each test case has a string S, which consists of only lowercase letters.

    The length of S is not larger than 10^5, and 1 ≤ M * L ≤ the length of S.
     
    Output
    For each test case, output the answer in a single line.
     
    Sample Input
    3 3
    abcabcbcaabc
     
    Sample Output
    2
    题目意思:问存在多少个子串,长度为n*l,而且这个子串由n个不同的 且长度都为l的串构成
    直接hash长度为l的串就可以了
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<string.h>
     4 #include<vector>
     5 #include<algorithm>
     6 #include<map>
     7 using namespace std;
     8 typedef unsigned long long ull;
     9 #define mmax 100000+10
    10 ull base[mmax],seed=131,ansx[mmax],ansy[mmax],aans[mmax];
    11 bool vit[mmax],vit2[mmax];
    12 map<ull,int>to;
    13 int n,m;
    14 char p[mmax];
    15 int main(){
    16     base[0]=1;
    17     for(int i=1;i<=100000;i++) base[i]=base[i-1]*seed;
    18     while(cin>>n>>m){
    19         scanf("%s",p);
    20         int len=strlen(p),tmp=n*m,id=0;
    21         memset(ansx,0,sizeof(ansx));
    22         for(int i=0;i<m;i++)
    23         ansx[0]=ansx[0]*seed+p[i];
    24         for(int i=m;i<len;i++) ansx[i-m+1]=ansx[i-m]*seed-p[i-m]*base[m]+p[i];
    25         memset(ansy,0,sizeof(ansy));
    26         for(int i=0;i<tmp;i++)
    27         ansy[0]=ansy[0]*seed+p[i];
    28         for(int i=tmp;i<len;i++) ansy[i-tmp+1]=ansy[i-tmp]*seed-p[i-tmp]*base[tmp]+p[i];
    29         memset(aans,0,sizeof(aans));
    30         for(int i=0;i<m;i++){
    31             to.clear();
    32             int first=i;
    33             for(int j=i,num=0;j<len+m;j+=m,num++){
    34                 if(num>=n){
    35                     if(to.size()==n) {
    36                         aans[id++]=ansy[first];
    37                     }
    38                     to[ansx[first]]--;
    39                     if(!to[ansx[first]]) to.erase(ansx[first]);
    40                     first+=m;
    41                 }
    42                 to[ansx[j]]++;
    43             }
    44         }
    45         sort(aans,aans+id);
    46 //        for(int i=0;i<id;i++) cout<<aans[i]<<" ";
    47 //        cout<<endl;
    48         if(id==0){
    49             cout<<0<<endl;
    50             continue;
    51         }
    52         int sum=0;
    53         for(int i=0;i<id;i++){
    54             if(aans[i]) {sum++;continue;}
    55 
    56         }
    57         cout<<sum<<endl;
    58     }
    59 }
     
  • 相关阅读:
    C++ primer学习方法
    windows 下安装使用ipython
    读书和思考
    win7 64位 python3.4&opencv3.0配置安装
    Deep Residual Learning for Image Recognition(MSRA-深度残差学习)
    MATLAB 常用形态学操作函数
    形态学图像处理
    限制对比度自适应直方图均衡(Contrast Limited Adaptive histgram equalization/CLAHE)
    计算机视觉,机器学习 ( 一些资源)
    对CNN模块的分析
  • 原文地址:https://www.cnblogs.com/ainixu1314/p/4227125.html
Copyright © 2011-2022 走看看