51nod1282(最小表示法&&枚举)

 1 #include <iostream>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 const int MAXN = 1e3 + 10;
 6 int a[MAXN][MAXN];
 7 
 8 void get_min(int n, int m){//最小表示法
 9     int i = 0, j = 1 ,k = 0, t;
10     while(i < m && j < m && k < m){
11         t = a[n][(i + k) % m] - a[n][(j + k) % m];
12         if (!t) k++;
13         else{
14             if(t > 0) i += k + 1;
15             else j += k + 1;
16             if (i == j) j++;
17             k = 0;
18         }
19     }
20     a[n][m] = i < j ? i : j;
21 }
22 
23 int main(void){
24     int n, m, p, ans = 0;
25     cin >> n >> m >> p;
26     for(int i = 0; i < n; i++){
27         for(int j = 0; j < m; j++){
28             cin >> a[i][j];
29         }
30         sort(a[i], a[i] + m);//注意给出的数据是未排序的
31         int x = a[i][0];
32         for(int j = 0; j < m - 1; j++){
33             a[i][j] = a[i][j + 1] - a[i][j];
34         }
35         a[i][m - 1] = x - a[i][m - 1] + p;//一开始没注意这里的a[i][0]变了，wa到死．．．
36         get_min(i, m);
37         int cnt = a[i][m], cc = m;
38         while(cc){
39             cnt = (cnt + 1) % m;
40             cc--;
41         }
42     }
43     for(int i = 0; i < n; i++){
44         for(int j = i + 1; j < n; j++){
45             int cnt = m, cnt1 = a[i][m], cnt2 = a[j][m];
46             while(a[i][cnt1] == a[j][cnt2] && cnt){
47                 cnt--;
48                 cnt1 = (cnt1 + 1) % m;
49                 cnt2 = (cnt2 + 1) % m;
50             }
51             if(!cnt) ans++;
52         }
53     }
54     cout << ans << endl;
55     return 0;
56 }