[Segment] zhongkui

Description
	Old manager is a smart and competent person. He has been working for the financial owner for 10 years, and the financial owner wants to make his account clearer. The manager is required to keep K accounts every day. Because the manager is smart and competent, the manager always makes the manager very satisfied. However, some people have doubts about the manager because of their provocation. So he decided to use a special method to judge the loyalty of the butler. He pressed every account to 1, 2, 3... Number, and then occasionally ask the manager's question, the question is: what is the smallest sum of Accounts A to B? In order to prevent the butler from committing frauds, he always asks multiple questions at a time.

	Input Format
		In the input, the first row has two numbers of M, N indicates M (M <= 100000), N indicates n problems, and n <= 100000. The number of M in the second action, which is the amount of money in the account. There are n problems in the next n rows. Each row has two numbers indicating the number of accounts starting to end.

	Output Format
		The output file contains the answer to each question. For details, see the example.

	Sample Input
		10 3 <br/> 1 2 3 4 5 6 7 8 9 10 <br/> 2 7 <br/> 3 9 <br/> 1 10

	Sample output sample output
		2 3 1

	Time Limit
		1 s for each test point

An entry into the segment.
The min value of each vertex is the min value of the left sub-shard and the min value of the right sub-shard.

Accode:

# Include <cstdio> # include <cstring> # include <cstdlib> # include <bitset> Using STD: min; const char fi [] = "tyvj1038.in "; const char fo [] = "tyvj1038.out"; const int maxn = 100010; const int maxm = 100010; const int max = 0x3fffff00; const int min =-Max; struct segtree {int L, R, LC, RC, Min ;}; segtree [maxm <1]; int T [maxn]; int n, m, TOT; void init_file () {freopen (FI, "r", stdin); freopen (FO, "W", St Dout);} void build (int l, int R) {int now = ++ tot; tree [now]. L = L; tree [now]. R = r; If (L = r) {tree [now]. min = T [l]; return;} int mid = (L + r)> 1; if (L <r) {tree [now]. lc = tot + 1; build (L, mid); tree [now]. rc = tot + 1; build (Mid + 1, R);} tree [now]. min = min (tree [tree [now]. LC]. min, tree [tree [now]. RC]. min);} int query (INT now, int L, int R) {If (L <= tree [now]. L & R> = tree [now]. r) Return tree [now]. min; int mid = (tree [now]. L + tree [now]. r)> 1; if (r <= mid) return query (tree [now]. LC, L, R); // subquestions that are completely included by the Left subaccount are directly related to the left subaccount. If (mid <L) return query (tree [now]. RC, L, R); // A subproblem that is directly related to the Right sub-region if the right sub-region contains it. Int lc = query (tree [now]. LC, L, R), Rc = query (tree [now]. RC, L, R); Return min (LC, RC);} void work () {scanf ("% d", & N, & M ); for (INT I = 1; I <n + 1; ++ I) scanf ("% d", T + I); Tot = 0; build (1, N ); for (; m; -- m) {int L, R; scanf ("% d", & L, & R); printf ("% d \ n ", query (1, L, R) ;}} int main () {init_file (); Work (); exit (0 );}