codeforces 1203A Circle of Students(思维)

思路很简单,用一个字符串保存原数组,对数组排序之后转成首尾相连的字符串,一个正序一个倒序再判断第一个字符串是不是他们中的子串就行了,这里顺便也练习了一下string的用法

#include<cstdio>
#include<stack>
#include<queue>
#include<cmath>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#define TP 233333333333333
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<char, int> P2;
const int maxn = 1000005;
int arr[maxn];
int main(void) {
    int n;
    cin >> n;
    while(n--) {
        int m;
        cin >> m;
        string s1, s2, rs;
        for (int i = 0; i<m; i++) {
            cin >> arr[i];
            s1 += (arr[i] + '0');
        }
        sort(arr, arr+m);
        for (int i = 0; i<m; i++) 
            s2 += (arr[i] + '0');
        s2 = s2 + s2;
        rs = s2;
        reverse(rs.begin(), rs.end());
        if (s2.find(s1) != string::npos || rs.find(s1) != string::npos)
            cout << "YES\n";
        else 
            cout << "NO\n";
    }
    return 0;
}
原文地址:https://www.cnblogs.com/shuitiangong/p/12266525.html