链接：
http://codeforces.com/contest/455/problem/D
题意：
给出一个序列，两种操作
1.a[l], a[l + 1], ..., a[r - 1], a[r] → a[r], a[l], a[l + 1], ..., a[r - 1].
2.在线询问区间[l,r]中值等于k的数有多少个
题解：
分块，每一块用双端队列表示就可以了，就是写的时候有很多细节要注意
代码：
31 int n, q; 32 int a[MAXN]; 33 int b[350][MAXN]; 34 deque<int> Block[350]; 35 int cap, num, L[MAXN], R[MAXN], Belong[MAXN]; 36 37 void init() { 38 cap = sqrt(n); 39 num = n / cap; 40 if (n%cap) num++; 41 rep(i, 1, n + 1) Belong[i] = (i - 1) / cap + 1; 42 rep(i, 1, num + 1) { 43 L[i] = (i - 1)*cap + 1; 44 R[i] = i*cap; 45 } 46 R[num] = n; 47 rep(i, 1, num + 1) { 48 rep(j, L[i], R[i] + 1) { 49 Block[i].pb(a[j]); 50 b[i][a[j]]++; 51 } 52 } 53 } 54 55 void update(int l, int r) { 56 if (Belong[l] == Belong[r]) { 57 int id = Belong[l]; 58 l -= L[id], r -= L[id]; 59 int t = Block[id][r]; 60 per(i, l + 1, r + 1) Block[id][i] = Block[id][i - 1]; 61 Block[id][l] = t; 62 return; 63 } 64 int id1 = Belong[l]; 65 int id2 = Belong[r]; 66 int t1 = Block[id2][r - L[id2]]; 67 int t2 = Block[id1][R[id1] - L[id1]]; 68 b[id1][t2]--; 69 b[id2][t1]--; 70 per(i, l + 1, R[id1] + 1) 71 Block[id1][i - L[id1]] = Block[id1][i - L[id1] - 1]; 72 Block[id1][l - L[id1]] = t1; 73 b[id1][t1]++; 74 rep(i, id1 + 1, id2) { 75 Block[i].push_front(t2); 76 b[i][t2]++; 77 t2 = Block[i].back(); 78 Block[i].pop_back(); 79 b[i][t2]--; 80 } 81 per(i, L[id2] + 1, r + 1) 82 Block[id2][i - L[id2]] = Block[id2][i - L[id2] - 1]; 83 Block[id2][0] = t2; 84 b[id2][t2]++; 85 } 86 87 int query(int l, int r, int k) { 88 int res = 0; 89 if (Belong[l] == Belong[r]) { 90 int id = Belong[l]; 91 rep(i, l, r + 1) if (Block[id][i - L[id]] == k) res++; 92 return res; 93 } 94 int id = Belong[l]; 95 rep(i, l, R[id] + 1) if (Block[id][i - L[id]] == k) res++; 96 rep(i, Belong[l] + 1, Belong[r]) res += b[i][k]; 97 id = Belong[r]; 98 rep(i, L[id], r + 1) if (Block[id][i - L[id]] == k) res++; 99 return res; 100 } 101 102 int main() { 103 ios::sync_with_stdio(false), cin.tie(0); 104 cin >> n; 105 rep(i, 1, n + 1) cin >> a[i]; 106 init(); 107 int ans = 0; 108 cin >> q; 109 while (q--) { 110 int op, l, r, k; 111 cin >> op >> l >> r; 112 l = (l + ans - 1) % n + 1; 113 r = (r + ans - 1) % n + 1; 114 if (l > r) swap(l, r); 115 if (op == 1) update(l, r); 116 else { 117 cin >> k; 118 k = (k + ans - 1) % n + 1; 119 ans = query(l, r, k); 120 cout << ans << endl; 121 } 122 } 123 return 0; 124 }