codeforce-191E-Thwarting Demonstrations(树状数组+二分+离散)

题意：

求第K 大连续区间

分析：

二分答案，再n * log(n)判断有几个区间的区间和大于mid，然后调整上下界，使这个值不断的接近k。

判断符合条件的区间总数：线性扫描sum[n](前n项和)  每次判断以i结尾的区间有几个区间和大于等于mid，累加即可

// File Name: 191-E.cpp
// Author: Zlbing
// Created Time: 2013年08月03日 星期六 15时10分13秒

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
const int MAXN=1e5+;
int tree[MAXN];
int n;
int lowbit(int x)
{
    return x&(-x);
}
void add(int pos,int val)
{
    while(pos<=n)
    {
        tree[pos]+=val;
        pos+=lowbit(pos);
    }
}
int read(int x)
{
    int s=;
    while(x>)
    {
        s+=tree[x];
        x-=lowbit(x);
    }
    return s;
}
LL sum[MAXN];
LL num[MAXN];
int N;
LL M;
LL solve(LL mid)
{
    CL(tree,);
    LL ans=;
    for(int i=;i<=N;i++)
    {
        if(sum[i]>=mid)ans++;
        LL t=sum[i]-mid;
        int a=upper_bound(num+,num+n+,t)-num-;
        int b=lower_bound(num+,num+n+,sum[i])-num;
        ans+=read(a);
        add(b,);
    }
    //cout<<"ans="<<ans<<endl;
    return ans;
}
int main()
{
    while(~scanf("%d%lld",&N,&M))
    {
        sum[]=;
        LL l,r;
        REP(i,,N)
        {
            scanf("%lld",&sum[i]);
            sum[i]+=sum[i-];
            num[i]=sum[i];
        }
        sort(num+,num+N+);
        n=unique(num+,num+N+)-num-;
            //cout<<"l="<<l<<"r="<<r<<endl;
        LL best=-;
        l=-1e18;
        r=1e18;
        while(l<=r)
        {
            LL mid=l+(r-l+)/;
            //cout<<"l="<<l<<"r="<<r<<"mid="<<mid<<endl;
            if(solve(mid)>=M)
            {
                best=mid;
                l=mid+;
            }
            else r=mid-;
        }
        cout<<best<<endl;
    }
    return ;
}