Ultraviolet A 11235-frequent values (rmq)

Ultraviolet A 11235-frequent values

Question Link

Given an ascending sequence, each time you ask for a range [L, R], you can find the maximum number of identical numbers.

Idea: rmq, because it is in ascending order, numbers of the same size must be connected to one piece. First, we can pre-process the total number of segments and the number of numbers contained in each segment, which segment corresponds to each position in the original array, the leftmost position and rightmost position. Then, you can divide the Query [L, R] into three segments each time you ask:
1. L to R [l] is a segment and the number is R [l]-l + 1.
2. L [R] To R is a segment, and the number is r-l [R] + 1
3. All the segments in the middle can be solved using rmq, with the number as the value.
The answer is a large number in the last three paragraphs.
Make sure that the selected [L, R] has only one and two segments.
Code:

# Include <cstdio >#include <cstring >#include <algorithm> using namespace STD; const int n = 100005; int N, Q, A [n], L [N], R [N], num [N], CNT [N], V, rmq [N] [20]; void Init () {v = 0; int count, pre; scanf ("% d", & Q); For (INT I = 0; I <n; I ++) {scanf ("% d ", & A [I]); If (! I | A [I]! = A [I-1]) {pre = I; if (I) CNT [V ++] = count; Count = 0;} count ++; num [I] = V; L [I] = pre;} CNT [V ++] = count; For (INT I = n-1; I> = 0; I --) {if (I = n-1 | A [I]! = A [I + 1]) {pre = I;} R [I] = pre;} void build_rmq (int * a, int N) {for (INT I = 0; I <n; I ++) rmq [I] [0] = A [I]; for (Int J = 1; (1 <j) <= N; j ++) {for (INT I = 0; I + (1 <j)-1 <n; I ++) {rmq [I] [J] = max (rmq [I] [J-1], rmq [I + (1 <(J-1)] [J-1]) ;}}} int query (int l, int R) {If (L> r) return 0; int K = 0; while (1 <(k + 1) <= r-L + 1) K ++; return max (rmq [l] [K], rmq [R-(1 <k) + 1] [k]);} Void solve () {build_rmq (CNT, V); int L, R; while (Q --) {scanf ("% d", & L, & R ); l --; r --; If (Num [l] = num [R]) printf ("% d \ n", R-l + 1 ); else printf ("% d \ n", max (query (Num [l] + 1, num [R]-1), max (R [l]-l + 1, r-l [R] + 1);} int main () {While (~ Scanf ("% d", & N) {Init (); solve ();} return 0 ;}