UVaOJ 1584 – Circular Sequence

问题描述

解法一：模拟

字典序（lexicographical order）概念题，可以自己实现字符串字典序的比较逻辑（这里选择调用 string.compare 方法）。另外需要实现一个循环左移 k 位操作（对于此题 k==1 ），可以当成是拆开成两个子串 s1+s2 的调换位置拼接，变成 s2+s1, 记得在多次左移操作的同时比较字典序维护 ans 的值。

#include <bits/stdc++.h>

using namespace std;

int main() {
  int t;
  cin >> t;
  while (t--) {
    string seq;
    cin >> seq;
    if (seq.length() == 1) {
      cout << seq << endl;
      continue;
    }

    string cur = seq, ans = seq;
    for (int i = 1; i < seq.length(); ++i) {
      cur = cur.substr(1, cur.length() - 1) + cur.substr(0, 1);
      if (cur.compare(ans) < 0) ans = cur;
    }
    cout << ans << endl;
  }
  return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main() {

int t;

cin >> t;

while (t--) {

string seq;

cin >> seq;

if (seq.length() == 1) {

cout << seq << endl;

continue;

}

string cur = seq, ans = seq;

for (int i = 1; i < seq.length(); ++i) {

cur = cur.substr(1, cur.length() - 1) + cur.substr(0, 1);

if (cur.compare(ans) < 0) ans = cur;

}

cout << ans << endl;

}

return 0;

}

t = int(input())
for _ in range(t):
    seq = input()
    if len(seq) == 1:
        print(seq)
        continue

    cur = ans = seq
    for i in range(1, len(seq)):
        cur = cur[1:len(cur)] + cur[:1]
        if cur < ans:
            ans = cur
    print(ans)

t = int(input())

for _ in range(t):

seq = input()

if len(seq) == 1:

print(seq)

continue

cur = ans = seq

for i in range(1, len(seq)):

cur = cur[1:len(cur)] + cur[:1]

if cur < ans:

ans = cur

print(ans)

AOAPC II UVaOJ 字符串

UVaOJ 1584 - Circular Sequence

问题描述

解法一：模拟