F. Isomorphic Strings
time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string s of length n consisting of lowercase English letters.

For two given strings s and t, say S is the set of distinct characters of s and T is the set of distinct characters of t. The strings s and t are isomorphic if their lengths are equal and there is a one-to-one mapping (bijection) f between S and T for which f(si) = ti. Formally:

  1. f(si) = ti for any index i,
  2. for any character  there is exactly one character  that f(x) = y,
  3. for any character  there is exactly one character  that f(x) = y.

For example, the strings "aababc" and "bbcbcz" are isomorphic. Also the strings "aaaww" and "wwwaa" are isomorphic. The following pairs of strings are not isomorphic: "aab" and "bbb", "test" and "best".

You have to handle m queries characterized by three integers x, y, len (1 ≤ x, y ≤ n - len + 1). For each query check if two substrings s[x... x + len - 1] and s[y... y + len - 1] are isomorphic.

Input

The first line contains two space-separated integers n and m (1 ≤ n ≤ 2·105, 1 ≤ m ≤ 2·105) — the length of the string s and the number of queries.

The second line contains string s consisting of n lowercase English letters.

The following m lines contain a single query on each line: xiyi and leni (1 ≤ xi, yi ≤ n, 1 ≤ leni ≤ n - max(xi, yi) + 1) — the description of the pair of the substrings to check.

Output

For each query in a separate line print "YES" if substrings s[xi... xi + leni - 1] and s[yi... yi + leni - 1] are isomorphic and "NO" otherwise.

Example
input
Copy
7 4
abacaba
1 1 1
1 4 2
2 1 3
2 4 3
output
Copy
YES
YES
NO
YES
Note

The queries in the example are following:

  1. substrings "a" and "a" are isomorphic: f(a) = a;
  2. substrings "ab" and "ca" are isomorphic: f(a) = cf(b) = a;
  3. substrings "bac" and "aba" are not isomorphic since f(b) and f(c) must be equal to a at same time;
  4. substrings "bac" and "cab" are isomorphic: f(b) = cf(a) = af(c) = b.

AC代码为:

#include<iostream>
#include<queue>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=2e5+10;
const ll MOD=1e9+7;
ll h[26][maxn],x=107,px[maxn];
char s[maxn];
int main()
{
    int n,m;
    scanf("%d%d", &n,&m);
    scanf("%s", s+1);
    px[0]=1;
    for(int i = 1; i <= n; ++i) px[i]=px[i-1]*x%MOD;
    for(int i = 0; i < 26; ++i)
    {
        for(int j = 1; j <= n; ++j) h[i][j]=(h[i][j-1]*x+int(s[j] == (i+'a')))%MOD;
    }
    while(m--)
    {
        int x,y,l;
        scanf("%d%d%d", &x,&y,&l);
        vector<int> p,q;
        for(int i = 0; i < 26; ++i)
        {
            p.push_back(((h[i][x+l-1]-px[l]*h[i][x-1]%MOD)%MOD+MOD)%MOD);
            q.push_back(((h[i][y+l-1]-px[l]*h[i][y-1]%MOD)%MOD+MOD)%MOD);
        }
        sort(q.begin(),q.end());sort(p.begin(),p.end());
        printf("%s\n", p==q? "YES":"NO");
    }
    return 0;
}

CodeForces985F -- Isomorphic Strings的更多相关文章

  1. [LeetCode] Isomorphic Strings

    Isomorphic Strings Total Accepted: 30898 Total Submissions: 120944 Difficulty: Easy Given two string ...

  2. leetcode:Isomorphic Strings

    Isomorphic Strings Given two strings s and t, determine if they are isomorphic. Two strings are isom ...

  3. Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings

    Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings 题目连接: http://cod ...

  4. [leetcode]205. Isomorphic Strings 同构字符串

    Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the chara ...

  5. Codeforces 985 F - Isomorphic Strings

    F - Isomorphic Strings 思路:字符串hash 对于每一个字母单独hash 对于一段区间,求出每个字母的hash值,然后排序,如果能匹配上,就说明在这段区间存在字母间的一一映射 代 ...

  6. Educational Codeforces Round 44 (Rated for Div. 2) F - Isomorphic Strings

    F - Isomorphic Strings 题目大意:给你一个长度为n 由小写字母组成的字符串,有m个询问, 每个询问给你两个区间, 问你xi,yi能不能形成映射关系. 思路:这个题意好难懂啊... ...

  7. LeetCode 205. 同构字符串(Isomorphic Strings)

    205. 同构字符串 205. Isomorphic Strings

  8. LeetCode_205. Isomorphic Strings

    205. Isomorphic Strings Easy Given two strings s and t, determine if they are isomorphic. Two string ...

  9. 【刷题-LeetCode】205. Isomorphic Strings

    Isomorphic Strings Given two strings *s* and *t*, determine if they are isomorphic. Two strings are ...

随机推荐

  1. 本地存储localstorage

    小小插件,封装了一个存取删 <script type="text/javascript"> /* *getItem(name) * *setItem(name,valu ...

  2. SqlServer2005 查询 第三讲 between

    在数据库的查询中最重要的是要知道命令的顺序,因为在sql命令中有许多的参数,例如distinct,top,in,order by,group by.......如果你不能理解什么时候该执行什么的话,很 ...

  3. 用正则表达式获取URL中的查询参数

    总结获取url中查询参数的两种方式 通过正则表达式获取单个参数 url中的所有查询参数可以通过 window.location.search 字段获取,以字符串的形式返回.并有固定的格式 ?param ...

  4. SpringBoot基本配置详解

    SpringBoot项目有一些基本的配置,比如启动图案(banner),比如默认配置文件application.properties,以及相关的默认配置项. 示例项目代码在:https://githu ...

  5. pat 1002 A+B for Polynomials (25 分)

    1002 A+B for Polynomials (25 分) This time, you are supposed to find A+B where A and B are two polyno ...

  6. 领扣(LeetCode)最长公共前缀 个人题解

    编写一个函数来查找字符串数组中的最长公共前缀. 如果不存在公共前缀,返回空字符串 "". 示例 1: 输入: ["flower","flow" ...

  7. 解决FirewallD is not running问题

    centos7 1.查看firewalld状态:systemctl status firewalld,如果是dead状态,即防火墙未开启. 2.开启防火墙systemctl start firewal ...

  8. Openlayers ol.interaction.Select取消默认选中效果

    说明: 在使用ol.interaction.Select进行点击查询时,默认会把点击选中的要素显示在地图上 我的需求是做轨迹回放,并可以点击轨迹上某一点,进行查询.这时候如果重新播放轨迹,会发现这个选 ...

  9. 菜鸟系列Fabric源码学习 — 区块同步

    Fabric 1.4 源码分析 区块同步 本文主要从源码层面介绍fabric peer同步区块过程,peer同步区块主要有2个过程: 1)peer组织的leader与orderer同步区块 2)pee ...

  10. JavaScript笔记九

    1.数组方法 reverse() - 可以用来反转一个数组,它会对原数组产生影响 concat() - 可以连接两个或多个数组,它不会影响原数组,而是新数组作为返回值返回 join() - 可以将一个 ...