• DNA Sorting--hdu1379


    DNA Sorting

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 2203    Accepted Submission(s): 1075

    Problem Description

     

    One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC'', this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG'' has only one inversion (E and D)--it is nearly sorted--while the sequence ``ZWQM'' has 6 inversions (it is as unsorted as can be--exactly the reverse of sorted).

    You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness'', from ``most sorted'' to ``least sorted''. All the strings are of the same length.


    This problem contains multiple test cases!

    The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

    The output format consists of N output blocks. There is a blank line between output blocks.

     

     

     

    Input

     

    The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (1 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.

     

     

     

    Output

     

    Output the list of input strings, arranged from ``most sorted'' to ``least sorted''. If two or more strings are equally sorted, list them in the same order they are in the input file.

     

     

     

    Sample Input

     

    1 10 6
    AACATGAAGG
    TTTTGGCCAA
    TTTGGCCAAA
    GATCAGATTT
    CCCGGGGGGA
    ATCGATGCAT

     

     

     

    Sample Output

     

    CCCGGGGGGA
    AACATGAAGG
    GATCAGATTT
    ATCGATGCAT
    TTTTGGCCAA
    TTTGGCCAAA

    这个题开始就没读懂,所以也不会做,后来学长讲完后才明白啥意思;

    就是比如第一行数据AACATGAAGG     首先定义sum=0, A大于它后面字符的个数为0,sum+=0,第二个同样,第三个C大于它后面字符的个数为3,sum+=3。。。以此类推!

    然后按每行数字从小到大输出字符串!!!

     

     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<algorithm>
     4 using namespace std;
     5 struct as 
     6 {
     7     char s[101];
     8     int m;
     9 }aa[200];
    10 bool cmp(as s,as t)
    11 {
    12     return s.m < t.m;
    13 }
    14 int main()
    15 {
    16     int n,i,j,k,t,a,b,q;
    17     scanf("%d",&n);
    18     while(n--)
    19     {
    20         scanf("%d%d",&a,&b);
    21         getchar();
    22         for(k=0;k<b;k++)
    23         {
    24             
    25             scanf("%s",aa[k].s);
    26             {
    27                 for(t=0;t<a;t++)
    28                 {
    29                     int sum=0;
    30                     for(i=0;i<a-1;i++)
    31                     {
    32                         for(j=i+1;j<a;j++)
    33                         if(aa[t].s[i]>aa[t].s[j])
    34                         sum++;
    35                     }
    36                     aa[t].m=sum;//将各组总数放在结构体中
    37                 }
    38                 
    39             }    
    40         }
    41         sort(aa, aa+b, cmp);//排序结构体
    42         for(i=0;i<b;i++)
    43         printf("%s
    ",aa[i].s);    
    44     }
    45     return 0;
    46 }

     

     

  • 相关阅读:
    Scala进阶之路-idea下进行spark编程
    Scala进阶之路-Spark本地模式搭建
    Scala进阶之路-Scala高级语法之隐式(implicit)详解
    Scala进阶之路-Spark底层通信小案例
    Scala进阶之路-并发编程模型Akka入门篇
    Scala进阶之路-统计商家id的标签数以及TopN示例案例分析
    Scala进阶之路-Scala中的泛型介绍
    Scala进阶之路-尾递归优化
    Scala进阶之路-Scala特征类与unapply反向抽取
    Java基础-爬虫实战之爬去校花网网站内容
  • 原文地址:https://www.cnblogs.com/Eric-keke/p/4679590.html
Copyright © 2020-2023  润新知