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  • codeforces 128B. String

    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    One day in the IT lesson Anna and Maria learned about the lexicographic order.

    String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), thatxi < yi, and for any j (1 ≤ j < ixj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages​​.

    The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.

    Input

    The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).

    Output

    Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).

    Sample test(s)
    input
    aa
    2
    
    output
    a
    
    input
    abc
    5
    
    output
    bc
    
    input
    abab
    7
    
    output
    b
    
    Note

    In the second sample before string "bc" follow strings "a", "ab", "abc", "b".

    题意是给你一个字符串,找到其第k小的连续子序列。我们可以用set或者优先队列模拟,一开始把所有的单个字符放进去,然后每次取出set中最小的字符串,然后把这个字符串加上其后面一个字符放入set中继续排序,连续k次。这题结构体里不能用char str[100000],因为会爆内存,要用string.

    #include<iostream>
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<math.h>
    #include<vector>
    #include<map>
    #include<set>
    #include<queue>
    #include<stack>
    #include<string>
    #include<algorithm>
    using namespace std;
    #define ll int
    #define inf 0x7fffffff
    #define maxn 100005
    char str[maxn],str1[maxn];
    string ss;
    struct node{
        string s;
        //char s[maxn];
        int idx;
    }a,b,temp,temp1,ans;
    
    bool operator<(node a,node b){
        return a.s<b.s;
    }
    
    multiset<node>myset;
    multiset<node>::iterator it;
    
    int main()
    {
        int n,m,i,j,len,tot;
        while(scanf("%s",str)!=EOF)
        {
            len=strlen(str);
            scanf("%d",&n);
            myset.clear();
            memset(str1,0,sizeof(str1));
            for(i=0;i<len;i++){
                str1[0]=str[i];
                a.s=str1;
                a.idx=i+1;
                myset.insert(a);
            }
            tot=0;
            while(!myset.empty()){
                it=myset.begin();
                temp=*it;
                myset.erase(it);
                tot++;
                if(tot==n){
                    ans=temp;
                    break;
                }
                if(temp.idx<=len-1){
                    ss=temp.s;
                    ss=ss+str[temp.idx];
                    temp1.s=ss;
                    temp1.idx=temp.idx+1;
                    myset.insert(temp1);
                }
    
            }
            if(tot==n){
                cout<<ans.s<<endl;
            }
            else printf("No such line.
    ");
        }
        return 0;
    }
    




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  • 原文地址:https://www.cnblogs.com/herumw/p/9464639.html
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