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 ;
- }
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