主席树水题 ....不带离散化 直接权值主席树 然后查询就ok了
#include <bits/stdc++.h>
const int MAXN=3e5+10;
using namespace std;
typedef struct node{
int l,r,sum;
}node;
node d[MAXN*21];int cnt;
int a[MAXN];
int rt[MAXN];
void update(int &x,int y,int l,int r,int t){
x=++cnt;d[x]=d[y];d[x].sum++;
if(l==r)return ;
int mid=(l+r)>>1;
if(t<=mid)update(d[x].l,d[y].l,l,mid,t);
else update(d[x].r,d[y].r,mid+1,r,t);
}
int ans,vul;
void querty(int x,int y,int l,int r){
if(l==r){ans=l;return ;}
int mid=(l+r)>>1;
if(d[d[y].l].sum-d[d[x].l].sum>vul)querty(d[x].l,d[y].l,l,mid);
if(d[d[y].r].sum-d[d[x].r].sum>vul)querty(d[x].r,d[y].r,mid+1,r);
}
int main(){
int n,m;scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)scanf("%d",&a[i]),update(rt[i],rt[i-1],1,m,a[i]);
int l,r,q;scanf("%d",&q);
for(int i=1;i<=q;i++){
scanf("%d%d",&l,&r);
ans=-1;vul=(r-l+1)/2;
//cout<<l<<" "<<r<<" "<<vul<<endl;
querty(rt[l-1],rt[r],1,m);
if(ans==-1)puts("no");
else printf("yes %d
",ans);
}
return 0;
}
2223: [Coci 2009]PATULJCI
Time Limit: 10 Sec Memory Limit: 259 MBSubmit: 1503 Solved: 663
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Description
Input
Output
10 3 1 2 1 2 1 2 3 2 3 3 8 1 2 1 3 1 4 1 5 2 5 2 6 6 9 7 10
Sample Input
no
yes 1
no
yes 1
no
yes 2
no
yes 3
yes 1
no
yes 1
no
yes 2
no
yes 3
Sample Output
HINT
Notice:输入第二个整数是序列中权值的范围Lim，即1<=ai(1<=i<=n)<=Lim。
1<=Lim<=10000