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  • TZOJ--3968: The K-th Substring (模拟)

    3968: The K-th Substring 

    时间限制(普通/Java):1000MS/3000MS     内存限制:65536KByte

    描述

    bdep__ gets a string of length N (1 ≤ N ≤ 100) today. And he needs its K-th (0 < K ≤ N*(N+1)/2) substring in alphabet order to solve a really hard problem. Obviously, a string of length N has N*(N+1)/2 substrings. For example, string "aba" has six substrings in alphabet order: "a", "a", "ab", "aba", "b", "ba".

    Unfortunately, it is Sunday now, which means bdep__ has so much boring homework to copy, especially that of Digital Circuits. As a good programmer you must be able to help bdep__ find out the K-th substring of a specified string rapidly so that he can "finish" his homework in time.

    输入

    Input will consist of multiple test cases. The first line contains an integer T (T ≤ 12) which is the number of test cases. Then for each test case:

    1. One line contains two integers N (the length of the string), K (indicates the K-th substring).
    2. One line contains a non-empty string that only contains small Latin letters ('a'-'z').

    输出

    For each case:

    • Output the K-th substring in alphabet order of the input string.

    样例输入

    2
    5 8
    trick
    4 5
    okay

    样例输出

    ri
    kay

    题目来源

    NKPC8

     

    题目链接:http://tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=3968

    题目大意:给一个N个字符的字符串,求出所有子串中按照字典序排序的第k个

    暴力模拟出所有子串,然后排序

     

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<algorithm> 
    using namespace std;
    int main()
    {
    	int t,i,j,x,y,k;
    	string a[10000],c;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%d %d",&x,&y);
    		cin>>c;
    		k=0;
    		for(i=0;i<x;i++)
    		{
    			for(j=1;j+i<=x;j++)
    			{
    				a[k++]=c.substr(i,j);
    			}
    		}
    		sort(a,a+k);
    		cout<<a[y-1]<<endl;
    	}
    }

     

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  • 原文地址:https://www.cnblogs.com/Anidlebrain/p/10029369.html
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