UVa1584

Circular Sequence

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status Practice UVA 1584 uDebug

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

题意：

有一个环状的字符串序列，仅包含四种字符----‘A’、‘T’、‘G’、‘C’。顺时针读取整个序列，需要你在这个环状序列中找到一个起点字符，从这个字符开始顺时针依次读取完整个字符串，使得得到的字符串的字典序最小。

输入：

情况数T，之后T行每行一个字符串表示任意起点字符开始顺时针读取的环形字符串，长度不小于2不大于100。

输出：

每种情况输出符合字典序最小的字符串解。

分析：

由于该题的字符串长度不大，所以我们可以通过列举字符串可能出现的所有情况找出字符串最小的解。列举的时候就是简单的模拟。其中使用两个函数带来方便：strcpy(s1,s2)将字符串s2复制到s1；strcmp(s1,s2)比较两字符串的字典序，小于0表示s1<s2,等于0表示s1=s2,大于0表示s1>s2。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 using namespace std;
 5 // A,C,G,T
 6 const int MAX_N = 100;
 7 char str[MAX_N + 10],mstr[MAX_N + 10];
 8 int main(){
 9     int T; cin >> T;
10     while(T--){
11         scanf("%s",str);
12         int len = strlen(str);
13         strcpy(mstr,str);
14         // 遍历比较
15         for(int i = 0 ; i < len ; i++){
16             char lastc = str[len - 1];
17             for(int j = len - 1 ; j >= 1 ; j--)
18                 str[j] = str[j - 1];
19             str[0] = lastc;
20             if(strcmp(str,mstr) < 0)
21                 strcpy(mstr,str);
22         }
23         printf("%s\n",mstr);
24     }
25     return 0;
26 }