题目描述

You are given an integer sequence of length N, a= {a1,a2,…,aN}, and an integer K.
a has N(N+1)⁄2 non-empty contiguous subsequences, {al,al+1,…,ar} (1≤l≤r≤N). Among them, how many have an arithmetic mean that is greater than or equal to K?

Constraints
All input values are integers.
1≤N≤2×10 5
1≤K≤10 9
1≤ai≤10 9

输入

Input is given from Standard Input in the following format:
N K
a1
a2
:
aN

输出

Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.

样例输入

3 6
7
5
7

样例输出

5

提示

All the non-empty contiguous subsequences of a are listed below:
{a1} = {7}
{a1,a2} = {7,5}
{a1,a2,a3} = {7,5,7}
{a2} = {5}
{a2,a3} = {5,7}
{a3} = {7}
Their means are 7, 6, 19⁄3, 5, 6 and 7, respectively, and five among them are 6 or greater. Note that {a1} and {a3} are indistinguishable by the values of their elements, but we count them individually.

 

首先对于所有数减去k，这样就不用除(r-l+1), 然后我们发现所求的就是有多少对l,r，使得sum[r]-sum[l-1] >= 0, sum是减去k之后的序列的前缀和

用树状数组对sum求有多少个顺序对，多加一个0这个数，代表从sum[r]-sum[0]对答案的贡献。

由于sum[i]可能很大，所以需要离散化 用unique

#include <bits/stdc++.h>
#define ll long long
const int mod=1e9+7;
const int maxn=2e5+7;
using namespace std;
//int s[maxn];
ll sum[maxn],t[maxn];
ll tree[maxn];
int nn,m;
void add(int x)
{
    while(x<=nn)//nn为上限，要大于离散化后的最大的数字，
    {
        tree[x]++;
        x+=x&-x;//向上更新，树状数组核心代码。此处就不解释了，不会的先学树状数组。
    }
}
ll query(int x)
{
    ll num=0;
    while(x)
    {
        num+=tree[x];
        x-=x&-x;//向下求和
    }
    return num;
}
int main()
{
    int n,k,a;
    scanf("%d %d",&n,&k);
    nn=n+10;
    for(int i=1;i<=n;i++){
        scanf("%d",&a);
        sum[i]=t[i]=sum[i-1]+a-k;
    }
    sort(t,t+n+1);
    m=unique(t,t+n+1)-t;
    for(int i=0;i<=n;i++){
        sum[i]=lower_bound(t,t+m+1,sum[i])-t+1;
    }
    ll ans=0;
    for(int i=0;i<=n;i++){
        ans+=query(sum[i]);
        add(sum[i]);
    }
    printf("%lld\n",ans);
    return 0;
}