Problem Description

Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.

Input

The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)

Output

For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.

Sample Input

1 10 10 0 5 2 7 5 4 3 8 7 7 2 8 6 3 5 0 1 3 1 1 9 4 0 1 0 3 5 5 5 5 1 4 6 3 1 5 7 5 7 3

Sample Output

Case 1: 4 0 0 3 1 2 0 1 5 1

Source

2012 ACM/ICPC Asia Regional Hangzhou Online

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给你n个数，让你在一段区间之内求不大于k的值得数有多少个

解决方法：

原模板不动，额外再家一个函数，用二分的方法看mid值去多少的时候《=所给的值，不断的更新ans，最后的ans一定是最大的值

下面代码

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;const int MAXN=100010;
int tree[30][MAXN];//表示每层每个位置的值
int sorted[MAXN];//已经排序的数
int toleft[30][MAXN];//toleft[p][i]表示第i层从1到i有多少个数分入左边void build(int l,int r,int dep)
{if(l==r)return;int mid=(l+r)>>1;int same=mid-l+1;//表示等于中间值而且被分入左边的个数for(int i=l;i<=r;i++)if(tree[dep][i]<sorted[mid])same--;int lpos=l;int rpos=mid+1;for(int i=l;i<=r;i++){if(tree[dep][i]<sorted[mid])//比中间的数小，分入左边tree[dep+1][lpos++]=tree[dep][i];else if(tree[dep][i]==sorted[mid]&&same>0){tree[dep+1][lpos++]=tree[dep][i];same--;}else  //比中间值大分入右边tree[dep+1][rpos++]=tree[dep][i];toleft[dep][i]=toleft[dep][l-1]+lpos-l;//从1到i放左边的个数
}build(l,mid,dep+1);build(mid+1,r,dep+1);}//查询区间第k大的数，[L,R]是大区间，[l,r]是要查询的小区间
int query(int L,int R,int l,int r,int dep,int k)
{if(l==r)return tree[dep][l];int mid=(L+R)>>1;int cnt=toleft[dep][r]-toleft[dep][l-1];//[l,r]中位于左边的个数if(cnt>=k){//L+要查询的区间前被放在左边的个数int newl=L+toleft[dep][l-1]-toleft[dep][L-1];//左端点加上查询区间会被放在左边的个数int newr=newl+cnt-1;return query(L,mid,newl,newr,dep+1,k);}else{int newr=r+toleft[dep][R]-toleft[dep][r];int newl=newr-(r-l-cnt);return query(mid+1,R,newl,newr,dep+1,k-cnt);}
}
int n,m;
int ans;
void solve(int s,int t,int k){int l=1;int r=t-s+1;ans=0;while(l<=r){int mid=(l+r)>>1;int temp=query(1,n,s,t,0,mid);if(temp<=k){ans=mid;l=mid+1;}elser=mid-1;}printf("%d\n",ans);
}int main(){int T;int cas=1;int s,t,k;scanf("%d",&T);while(T--){memset(tree,0,sizeof(tree));scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){scanf("%d",&tree[0][i]);sorted[i]=tree[0][i];}sort(sorted+1,sorted+n+1);build(1,n,0);printf("Case %d:\n",cas++);for(int i=1;i<=m;i++){scanf("%d%d%d",&s,&t,&k);s++;t++;solve(s,t,k);}}return 0;
}