1584 - Circular Sequence

Circular Sequence

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

长度为n的环状串有n种表示法，分别为从某个位置开始顺时针得到。例如，图3-4的环状串有10种表示：

CGAGTCAGCT，GAGTCAGCTC，AGTCAGCTCG等。在这些表示法中，字典序最小的称为”最小表示”。

输入一个长度为n（n≤100）的环状DNA串（只包含A、C、G、T这4种字符）的一种表示法，你的任务是输出该环状串的最小表示。例如，CTCC的最小表示是CCCT，CGAGTCAGCT的最小表示为AGCTCGAGTC。

#include <stdio.h>#include <string.h>#define maxNum 105// 判断以位置p开头的小还是以位置q开头有的小int comp(const char* s, int p, int q) {    int n = strlen(s);    for(int i = 0; i < n; i++) {        if(s[(p + i) % n] != s[(q + i) % n]) {            // 位置q小于位置p返回true            // 大于返回false            return s[(p + i) % n] < s[(q + i) % n];        }    }    // 相等情况    return 0;}int main() {    int T,N;    scanf("%d", &T);    while(T--) {        char s[maxNum];        scanf("%s", s);        int n = strlen(s);        // 记录最小字典序开始位置        int pos = 0;        for(int i = 0; i < n; i++) {            if(comp(s, i, pos)) {                pos = i;            }        }        for(int i = 0; i < n; i++) {            putchar(s[(pos + i) % n]);        }        putchar('\n');    }    return 0;}