zoukankan      html  css  js  c++  java
  • kmp(前缀出现次数next应用)

    http://acm.hdu.edu.cn/showproblem.php?pid=3336

    Count the string

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 17068    Accepted Submission(s): 7721


    Problem Description
    It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:
    s: "abab"
    The prefixes are: "a", "ab", "aba", "abab"
    For each prefix, we can count the times it matches in s. So we can see that prefix "a" matches twice, "ab" matches twice too, "aba" matches once, and "abab" matches once. Now you are asked to calculate the sum of the match times for all the prefixes. For "abab", it is 2 + 2 + 1 + 1 = 6.
    The answer may be very large, so output the answer mod 10007.
     
    Input
    The first line is a single integer T, indicating the number of test cases.
    For each case, the first line is an integer n (1 <= n <= 200000), which is the length of string s. A line follows giving the string s. The characters in the strings are all lower-case letters.
     
    Output
    For each case, output only one number: the sum of the match times for all the prefixes of s mod 10007.
     
    Sample Input
    1 4 abab
     
    Sample Output
    6
     
    Author
    foreverlin@HNU
     
    Source
     
    Recommend
    lcy
    题意:求字符串所有前缀出现的次数(包括本身)
    思路:求所有next的值的个数加上前缀本身
     
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <iostream>
    #include <algorithm>
    #include <iostream>
    #include<cstdio>
    #include<string>
    #include<cstring>
    #include <stdio.h>
    #include <string.h>
    using namespace std;
    char a[200009];
    
    int num;
    
    
    void getnext(char* a, int len , int *next)
    {
        next[0] = -1 ;
        int k = -1 , j = 0 ;
        while(j < len)
        {
            if(k == -1 || a[j] == a[k])
            {
                k++;
                j++;
            //    if(a[j] != a[k])
                    next[j] = k ;
              //  else
              //  {
              //      next[j] = next[k];
               // }
    
            }
            else
            {
                k = next[k];
            }
        }
    }
    
    int main()
    {int n ;
        scanf("%d" , &n);
        while(n--)
        {
            int next[200009];
            int l ;
            scanf("%d" , &l);
            scanf("%s" , a);
            getnext(a , l , next);
            int j = 0 ;
            for(int i = 0 ; i <= l ; i++)
            {
            //    cout << next[i] << " " ;
                j = i ;
                while(next[j] > 0)
                {
                    num = (num + 1) % 10007 ;
                    j = next[j];
                }
            }
           // cout << endl ;
    
            printf("%d
    " , (num + l)%10007);
            num = 0 ;
        }
    
        return 0 ;
    }
  • 相关阅读:
    量化平台的发展转
    jmeter全面总结8jmeter实战
    月见笔谈【一】——关于悲剧
    为什么要不断接触和学习新技术之我见
    WPF后台动态调用样式文件
    WPF后台动态添加TabItem并设置样式
    SQL查询SQLSERVER数据库中的临时表结构脚本
    防抖功能的实现
    项目中自定义进度条的实现
    vue3 请求响应拦截
  • 原文地址:https://www.cnblogs.com/nonames/p/11296242.html
Copyright © 2011-2022 走看看