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
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.
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!!!
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct as
{
char s[];
int m;
}aa[];
bool cmp(as s,as t)
{
return s.m < t.m;
}
int main()
{
int n,i,j,k,t,a,b,q;
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&a,&b);
getchar();
for(k=;k<b;k++)
{ scanf("%s",aa[k].s);
{
for(t=;t<a;t++)
{
int sum=;
for(i=;i<a-;i++)
{
for(j=i+;j<a;j++)
if(aa[t].s[i]>aa[t].s[j])
sum++;
}
aa[t].m=sum;//将各组总数放在结构体中
} }
}
sort(aa, aa+b, cmp);//排序结构体
for(i=;i<b;i++)
printf("%s\n",aa[i].s);
}
return ;
}
DNA Sorting--hdu1379的更多相关文章
- 算法:POJ1007 DNA sorting
这题比较简单,重点应该在如何减少循环次数. package practice; import java.io.BufferedInputStream; import java.util.Map; im ...
- poj 1007:DNA Sorting(水题,字符串逆序数排序)
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 80832 Accepted: 32533 Des ...
- [POJ1007]DNA Sorting
[POJ1007]DNA Sorting 试题描述 One measure of ``unsortedness'' in a sequence is the number of pairs of en ...
- DNA Sorting 分类: POJ 2015-06-23 20:24 9人阅读 评论(0) 收藏
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 88690 Accepted: 35644 Descrip ...
- poj 1007 (nyoj 160) DNA Sorting
点击打开链接 DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 75164 Accepted: 30 ...
- [POJ] #1007# DNA Sorting : 桶排序
一. 题目 DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 95052 Accepted: 382 ...
- poj 1007 DNA Sorting
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 95437 Accepted: 38399 Des ...
- DNA Sorting(排序)
欢迎参加——BestCoder周年纪念赛(高质量题目+多重奖励) DNA Sorting Time Limit: 2000/1000 MS (Java/Others) Memory Limit: ...
- [POJ 1007] DNA Sorting C++解题
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 77786 Accepted: 31201 ...
- HDU 1379:DNA Sorting
DNA Sorting Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Tota ...
随机推荐
- 解决ScrollView中嵌套ListView滚动效果冲突问题
在ScrollView中嵌套使用ListView,ListView只会显示一行到两行的数据.起初我以为是样式的问题,一直在对XML文件的样 式进行尝试性设置,但始终得不到想要的效果.后来在网上查了查, ...
- Python 之socket的应用
本节主要讲解socket编程的有关知识点,顺便也会讲解一些其它的关联性知识: 一.概述(socket.socketserver): python对于socket编程,提供了两个模块,分别是socket ...
- project euler 26:Reciprocal cycles
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with d ...
- 我是菜鸟,我怕谁--hdu2520
我是菜鸟,我怕谁 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Su ...
- Mysql服务启动问题
mysql ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/tmp/mysql.sock' (2) u ...
- JavaEE Tutorials (30) - Duke综合案例研究示例
30.1Duke综合应用的设计和架构456 30.1.1events工程458 30.1.2entities工程459 30.1.3dukes—payment工程461 30.1.4dukes—res ...
- jQuery UI 之 EasyUI 快速入门
jQuery EasyUI 基础 转载自(http://www.shouce.ren/api/view/a/3350) jQuery EasyUI 是一个基于 jQuery 的框架,集成了各种用户界面 ...
- Jquery EasyUI修改行背景的两种方式
1.数据加载完成不请求后台的做法 方式一: //更改表格行背景 function changeLineStyle(index){ var rows=$("#alertGird"). ...
- UESTC_秋实大哥掰手指 2015 UESTC Training for Dynamic Programming<Problem B>
B - 秋实大哥掰手指 Time Limit: 3000/1000MS (Java/Others) Memory Limit: 2048/1024KB (Java/Others) Submit ...
- 操作系统基本概念(内核态与用户态、操作系统结构)-by sixleaves
内核态与用户态(为什么存在这种机制.程序应处于哪个状态.如何判断当前所处状态.哪些功能需要内核态.如何实现这种机制) 1.首先我们应该思考清楚为什么会有内核态和用户态?(为什么存在这种机制) 因为计算 ...