CF 999 B. Reversing Encryption

题目地址

分析:

  规律:当i为n的倍数时,从0至i翻转(从前往后)

       新知识: reverse(s+a,s+b);把字符串s中,地址从a到b间的内容反转 

代码:

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstring>
 4 using namespace std;
 5 
 6 int main()
 7 {
 8     char s[105];
 9     int n;
10     cin >> n >> s; 
11     for(int i = 1; i <= n; i++)
12         if(n%i == 0)
13             reverse(s,s+i);
14     cout << s << endl;
15     return 0;
16 }

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转载自www.cnblogs.com/JiaaaaKe/p/9502495.html