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。
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。
m行，每行对应一个答案。
7 5
1 1 3 2 3 4 3
1 3
1 4
3 7
1 7
6 6
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
");
else printf("%d
", b[t]);
}
return 0;
}