CodeForces - 798B Mike and strings

B. Mike and strings
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

Mike has n strings s1, s2, …, sn each consisting of lowercase English letters. In one move he can choose a string si, erase the first character and append it to the end of the string. For example, if he has the string “coolmike”, in one move he can transform it into the string “oolmikec”.

Now Mike asks himself: what is minimal number of moves that he needs to do in order to make all the strings equal?

Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of strings.

This is followed by n lines which contain a string each. The i-th line corresponding to string si. Lengths of strings are equal. Lengths of each string is positive and don’t exceed 50.

Output
Print the minimal number of moves Mike needs in order to make all the strings equal or print - 1 if there is no solution.

Examples
input
4
xzzwo
zwoxz
zzwox
xzzwo
output
5
input
2
molzv
lzvmo
output
2
input
3
kc
kc
kc
output
0
input
3
aa
aa
ab
output
-1
Note
In the first sample testcase the optimal scenario is to perform operations in such a way as to transform all strings into “zwoxz”.

问题链接：CodeForces - 798B Mike and strings
问题简述：（略）
问题分析：字符串循环比较问题，按理说应该用字符串循环比较来实现，时间上会比较快。字符串循环比较可以通过模除计算来实现。一个简易的处理是把同一字符串连起来替代字符串比较，用string类来处理是比较方便的。
把这个程序的string方法（函数）find()，用自己编写的函数来实现，可以做出一个高效的程序。
AC的C++语言程序如下：

/* CodeForces - 798B Mike and strings */#include <bits/stdc++.h>using namespace std;const int INF = 0x3F3F3F3F;
const int N = 50;
string s[N];int find(string &a, string &b)
{int len = a.length();for (int i = 0; i < len; i++) {int j;for (j = 0; j < b.length(); j++)if(a[(i + j) % len] != b[j]) break;if(j == b.length()) return i;}return -1;
}int main()
{int n;cin >> n;for (int i = 0; i < n; i++) cin >> s[i];int flag = 1, ans = INF;for (int i = 0; i < n && flag; i++) {int sum = 0, pos;for (int j = 0; j < n; j++) {if ((pos = find(s[j], s[i])) == -1) {flag = 0;break;} elsesum += pos;}if (flag) ans = min(ans, sum);}if (flag) cout << ans << endl;else cout << -1 << endl;return 0;
}

AC的C++语言程序如下：

/* CodeForces - 798B Mike and strings */#include <bits/stdc++.h>using namespace std;const int INF = 0x3F3F3F3F;
const int N = 50;
string s[N], s2[N];int main()
{int n;cin >> n;for (int i = 0; i < n; i++) {cin >> s[i];s2[i] = s[i] + s[i];}int flag = 1, ans = INF;for (int i = 0; i < n && flag; i++) {int sum = 0, pos;for (int j = 0; j < n; j++)if ((pos = s2[j].find(s[i])) == string::npos) {flag = 0;break;} elsesum += pos;if (flag) ans = min(ans, sum);}if (flag) cout << ans << endl;else cout << -1 << endl;return 0;
}

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