BZOJ5178: [Jsoi2011]棒棒糖

BZOJ

题意

给你(n)个数,每次询问一个区间([l,r])是否存在一个出现次数严格大于(frac{r-l+1}{2})的数,如果有,输出这个数,否则输出0;

题解

主席树上二分,到了叶子节点再判一下出现次数;

#include<bits/stdc++.h>
#define Fst first
#define Snd second
#define RG register
#define mp make_pair
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned int UI;
typedef unsigned long long ULL;
template<typename T> inline void read(T& x) {
	char c = getchar();
	bool f = false;
	for (x = 0; !isdigit(c); c = getchar()) {
		if (c == '-') {
			f = true;
		}
	}
	for (; isdigit(c); c = getchar()) {
		x = x * 10 + c - '0';
	}
	if (f) {
		x = -x;
	}
}
template<typename T, typename... U> inline void read(T& x, U& ... y) {
	read(x), read(y...);
}
const int MAX=5e4,N=MAX+10;
int n,Q,size;
int root[N];
struct Node {
	int lo,ro,sum;
}Tr[N*20];
void Modify(int l,int r,int &o,int pos) {
	Tr[++size]=Tr[o]; o=size;
	++Tr[o].sum;
	if(l==r) return;
	int mid=l+r>>1;
	if(pos<=mid) Modify(l,mid,Tr[o].lo,pos);
	else Modify(mid+1,r,Tr[o].ro,pos);
}
int Query(int l,int r,int x,int y,int len,int g) {
	if(l==r) return Tr[x].sum-Tr[y].sum>len?l:0;
	int mid=l+r>>1,t=Tr[Tr[x].lo].sum-Tr[Tr[y].lo].sum;
	if(t+g>len) return Query(l,mid,Tr[x].lo,Tr[y].lo,len,g);
	return Query(mid+1,r,Tr[x].ro,Tr[y].ro,len,g+t);
}
int main() {
//	ios::sync_with_stdio(false);
#ifdef rua
	freopen("GG.in","r",stdin);
#endif
	read(n,Q);
	for(int i=1;i<=n;++i) {
		int a; read(a);
		Modify(1,MAX,root[i]=root[i-1],a);
	}
	while(Q--) {
		int l,r; read(l,r);
		printf("%d
",Query(1,MAX,root[r],root[l-1],(r-l+1)/2,0));
	}
	return 0;
}