[POI2014]KUR-Couriers BZOJ3524 主席树

给一个长度为n的序列a。1≤a[i]≤n。
m组询问，每次询问一个区间[l,r]，是否存在一个数在[l,r]中出现的次数大于(r-l+1)/2。如果存在，输出这个数，否则输出0。

Input

第一行两个数n，m。
第二行n个数，a[i]。
接下来m行，每行两个数l,r，表示询问[l,r]这个区间。

Output

m行，每行对应一个答案。

Sample Input

7 5
1 1 3 2 3 4 3
1 3
1 4
3 7
1 7
6 6

Sample Output

1
0
3
0
4

仍然是基础的主席树；
不离散化也可以；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#pragma GCC optimize(2)
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 700005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)

inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/



ll qpow(ll a, ll b, ll c) {
	ll ans = 1;
	a = a % c;
	while (b) {
		if (b % 2)ans = ans * a%c;
		b /= 2; a = a * a%c;
	}
	return ans;
}

int n, m, q;
int cnt = 0;
int a[maxn], b[maxn], T[maxn];
int sum[maxn << 5], lson[maxn << 5], rson[maxn << 5];

int build(int l, int r) {
	int rt = ++cnt;
	sum[rt] = 0;
	if (l < r) {
		int mid = (l + r) >> 1;
		lson[rt] = build(l, mid); rson[rt] = build(mid + 1, r);
	}
	return rt;
}

int  upd(int pre, int l, int r, int  x) {
	int rt = ++cnt;
	lson[rt] = lson[pre]; rson[rt] = rson[pre];
	sum[rt] = sum[pre] + 1;
	if (l < r) {
		int mid = (l + r) >> 1;
		if (x <= mid)lson[rt] = upd(lson[pre], l, mid, x);
		else rson[rt] = upd(rson[pre], mid + 1, r, x);
	}
	return rt;
}

int query(int u, int v, int l, int r, int k) {
	if (l >= r)return l;
	int x = sum[lson[v]] - sum[lson[u]];
	int y = sum[rson[v]] - sum[rson[u]];
	int mid = (l + r) >> 1;
	if (x > k)return query(lson[u], lson[v], l, mid, k);
	else if (y > k) return query(rson[u], rson[v], mid + 1, r, k);
	else return 0;
}

int main(){
	//ios::sync_with_stdio(0);
	rdint(n); rdint(q);
	for (int i = 1; i <= n; i++)rdint(a[i]), b[i] = a[i];
	sort(b + 1, b + 1 + n);
	m = unique(b + 1, b + 1 + n) - b - 1;
	T[0] = build(1, m);
	for (int i = 1; i <= n; i++) {
		int t = lower_bound(b + 1, b + 1 + m, a[i]) - b;
		T[i] = upd(T[i - 1], 1, m, t);
	}
	while (q--) {
		int x, y, z; rdint(x); rdint(y); z = (y - x + 1) >> 1;
		int t = query(T[x - 1], T[y], 1, m, z);
		if (t == 0)printf("0\n");
		else printf("%d\n", b[t]);
	}
    return 0;
}