The hh-index of an author is the largest hh where he has at least hh papers with citations not less than hh.

Bobo has published nn papers with citations a1,a2,…,ana1,a2,…,an respectively.
One day, he raises qq questions. The ii-th question is described by two integers lili and riri, asking the hh-index of Bobo if has only published papers with citations ali,ali+1,…,ariali,ali+1,…,ari.
Input
The input consists of several test cases and is terminated by end-of-file.

The first line of each test case contains two integers nn and qq.
The second line contains nn integers a1,a2,…,ana1,a2,…,an.
The ii-th of last qq lines contains two integers lili and riri.
Output
For each question, print an integer which denotes the answer.

Constraint

• 1≤n,q≤1051≤n,q≤105
• 1≤ai≤n1≤ai≤n
• 1≤li≤ri≤n1≤li≤ri≤n
• The sum of nn does not exceed 250,000250,000.
• The sum of qq does not exceed 250,000250,000.
  Sample Input
  5 3
  1 5 3 2 1
  1 3
  2 4
  1 5
  5 1
  1 2 3 4 5
  1 5
  Sample Output
  2
  2
  2
  3
  題意：一組長度為n的序列，m次查詢。每次查詢給出兩個數(shù)字l和r。詢問的是找出最大的數(shù)字k，使得l到r之間的數(shù)字至少有k個數(shù)字大于等于k。
  思路很簡單，二分答案，主席樹上求區(qū)間大于等于k的數(shù)字個數(shù)。
  很多題解說的是求區(qū)間第k大，其實沒有那么麻煩，直接離散化之后二分答案k，求大于等于k的個數(shù)就行。根據(jù)個數(shù)縮短區(qū)間。
  代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=1e5+100; struct node{int l;int r;int sum; }p[maxx*40]; int a[maxx],b[maxx],root[maxx]; int n,m,tot; /*--------------主席樹--------------*/ inline int build(int l,int r) {int cur=++tot;p[cur].sum=0;if(l==r) return cur;int mid=l+r>>1;p[cur].l=build(l,mid);p[cur].r=build(mid+1,r);return cur; } inline int update(int rot,int pos,int l,int r) {int cur=++tot;p[cur]=p[rot];p[cur].sum++;if(l==r) return cur;int mid=l+r>>1;if(pos<=mid) p[cur].l=update(p[rot].l,pos,l,mid);else p[cur].r=update(p[rot].r,pos,mid+1,r);return cur; } inline int query(int lrot,int rrot,int l,int r,int pos) {if(l==r) return p[rrot].sum-p[lrot].sum;int mid=l+r>>1;int sum=0;if(pos<=mid) sum=query(p[lrot].l,p[rrot].l,l,mid,pos)+p[p[rrot].r].sum-p[p[lrot].r].sum;else sum=query(p[lrot].r,p[rrot].r,mid+1,r,pos);return sum; } int main() {int l,r;while(~scanf("%d%d",&n,&m)){tot=0;for(int i=1;i<=n;i++) scanf("%d",&a[i]),b[i]=a[i];sort(b+1,b+1+n);int len=unique(b+1,b+1+n)-b-1;//離散化root[0]=build(1,len);for(int i=1;i<=n;i++) root[i]=update(root[i-1],lower_bound(b+1,b+1+len,a[i])-b,1,n);while(m--){scanf("%d%d",&l,&r);int L=1,R=n,mid,ans;while(L<=R){mid=L+R>>1;int zz=query(root[l-1],root[r],1,n,lower_bound(b+1,b+1+len,mid)-b);if(zz>=mid){ans=mid;L=mid+1;}else R=mid-1;}printf("%d\n",ans);}}return 0; }

努力加油a啊，(^o)/~

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