UVA 1584 Circular Sequence

Circular Sequence

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description





Some DNA sequences exist in circular forms as in the following figure, which shows a circular sequence ``CGAGTCAGCT", that is, the last symbol ``T" in ``CGAGTCAGCT" is connected to the first symbol ``C". We always read a circular
sequence in the clockwise direction.



Since it is not easy to store a circular sequence in a computer as it is, we decided to store it as a linear sequence. However, there can be many linear sequences that are obtained from a circular sequence by cutting any place
of the circular sequence. Hence, we also decided to store the linear sequence that is lexicographically smallest among all linear sequences that can be obtained from a circular sequence.
Your task is to find the lexicographically smallest sequence from a given circular sequence. For the example in the figure, the lexicographically smallest sequence is ``AGCTCGAGTC". If there are two or more linear sequences that
are lexicographically smallest, you are to find any one of them (in fact, they are the same).

Input

The input consists of T test cases. The number of test cases T is given on the first line of the input file.
Each test case takes one line containing a circular sequence that is written as an arbitrary linear sequence. Since the circular sequences are DNA sequences, only four symbols, A, C, G and T,
are allowed. Each sequence has length at least 2 and at most 100.

Output

Print exactly one line for each test case. The line is to contain the lexicographically smallest sequence for the test case.
The following shows sample input and output for two test cases.

Sample
Input

2
CGAGTCAGCT
CTCC

Sample
Output

AGCTCGAGTC
CCCT

注意：

1：尽量用define 定义

2：表示方法相同

int less(char *s,int a,int b)

int less(char s[],int a,int b)

3：环状数组如33行表示

#include<stdio.h>
#include<string.h>
#define maxn 105
//环状S的表示法a是否比表示法b字典序小
//int less(char  *s,int a,int b) //表示方法相同
int less(char s[],int a,int b)
{
int n=strlen(s);
for(int i=0;i<n;i++)
{
if(s[(i+a)%n]!=s[(i+b)%n])
{
return s[(i+a)%n]<s[(i+b)%n];
}
}
return 0;//相等
}
int main (void)
{
int T;
char s[maxn];
scanf("%d",&T);
while(T--&&scanf("%s",s))
{
int n=strlen(s);
int sec=0;
for(int i=1;i<n;i++)
{
if(less(s,i,sec)) sec=i;
}
for(int i=0;i<n;i++)
{
putchar(s[(i+sec)%n]);//%n长度 变成环状
}
printf("\n");
}
return 0;
}