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
- TCP连接状态图
- 使用jekyll和prose在github上创建博客
利用github的pages服务可以很方便地显示和管理我们的静态页面,这样用来做博客是非常适合的. 1.首先你要有一个github的帐号 2.创建一个repo,名字叫username.github.i ...
- linux设置虚拟内存(swap)解决mysql因内存不足挂掉的故障
mysql错误日志显示: InnoDB: mmap(137363456 bytes) failed; errno 122016-03-01 01:38:42 13064 [ERROR] InnoDB: ...
- js的eval方法
eval方法: 检查JScript代码,并且执行. 语法: eval(codeString); 参数: codeString是必选项,参数是包含有效的JScript代码的字符串值,这个字符串值将由JS ...
- Js函数加括号、不加括号(转)
函数只要是要调用它进行执行的,都必须加括号.此时,函数()实际上等于函数的返回值.当然,有些没有返回值,但已经执行了函数体内的行为,这个是根本,就是说,只要加括号的,就代表将会执行函数体代码. 不加括 ...
- 华为IC设计人员的薪酬(5年经验28万),以及麒麟是如何脱颖而出的~
垂直整合助力麒麟腾飞 由于ARM技术路线大幅降低了技术门槛和研发的资金和时间成本,导致ARM阵营参与者众多,加上大家都是购买ARM的CPU核与GPU核,造成产品高度同质化,市场竞争异常激烈——在价格上 ...
- VC++6.0出现no compile tool is associated with the extension.解决方法
对于刚解除VC++6.0的小白,在编译时候经常出现下图的错误提示: 解释为:不能编译此BmpRot.h文件,没有合适的编译工具可以编译此扩展名的文件. 很明显,当然只有.cpp文件才能编译. .h头文 ...
- java Socket使用注意
Socket s = new Socket(ia, port); BufferedOutputStream bufOut = new BufferedOutputStream(s.getOutputS ...
- Unity 调用android插件
1. Unity的Bundle Identifier必须和你的android报名一致 Activity和View的区别: Activity应该是一个展示页面,View是页面上一些按钮视图等等. 如何调 ...
- eclipse默认编码设置为utf-8
需要设置的几处地方为: Window->Preferences->General ->Content Type->Text->JSP 最下面设置为UTF-8 Window ...