Petya and Array (权值线段树+逆序对)

最新推荐文章于 2024-12-12 19:33:59 发布

转载最新推荐文章于 2024-12-12 19:33:59 发布 · 175 阅读

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原文链接：http://www.cnblogs.com/Fighting-sh/p/9719751.html

本文介绍了一种高效算法，用于解决给定数组中区间和小于特定阈值t的个数问题。通过使用前缀和与逆序对的概念，结合自定义的数据结构和算法优化，实现了快速求解。文章提供了详细的代码实现与示例，帮助读者理解并应用此算法。

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Petya and Array

http://codeforces.com/problemset/problem/1042/D

Petya has an array $aa consisting of nn integers. He has learned partial sums recently, and now he can calculate the sum of elements on any segment of the array really fast. The segment is a non-empty sequence of elements standing one next to another in the array.$

Now he wonders what is the number of segments in his array with the sum less than $tt. Help Petya to calculate this number.$

More formally, you are required to calculate the number of pairs $l,rl,r (l\leqrl\leqr) such that al+al+1+\dots+ar-1+ar<tal+al+1+\dots+ar-1+ar<t.$

Input

The first line contains two integers $nn and tt (1\leqn\leq200000,|t|\leq2\cdot10141\leqn\leq200000,|t|\leq2\cdot1014).$

The second line contains a sequence of integers ${a1,a2,\dots,ana1,a2,\dots,an (|ai|\leq109|ai|\leq109) — the description of Petya's array. Note that there might be negative, zero and positive elements.}_{1,a2,\dots,ana1,a2,\dots,an (|ai|\leq109|ai|\leq109) — the description of Petya's array. Note that there might be negative, zero and positive elements.}$

Output

Print the number of segments in Petya's array with the sum of elements less than $tt.$

Examples

input

5 4
5 -1 3 4 -1

output

input

3 0
-1 2 -3

output

input

4 -1
-2 1 -2 3

output

Note

In the first example the following segments have sum less than $44:$

$[2,2][2,2], sum of elements is -1-1$
$[2,3][2,3], sum of elements is 22$
$[3,3][3,3], sum of elements is 33$
$[4,5][4,5], sum of elements is 33$
$[5,5][5,5], sum of elements is -1$

$参考博客 http://mamicode.com/info-detail-2452129.html$

　　找区间和小于t的个数，区间和的问题，一般用前缀和来做

　　可以看成sum[i]>t+sum[k]的个数，i<k<=n。这样就变成了一个逆序对的问题

  1 #include<iostream>
  2 #include<cstring>
  3 #include<string>
  4 #include<cmath>
  5 #include<algorithm>
  6 #include<queue>
  7 #include<cstdio>
  8 #include<vector>
  9 #define maxn 500005
 10 #define lson l,mid,rt<<1
 11 #define rson mid+1,r,rt<<1|1
 12 typedef long long ll;
 13 using namespace std;
 14 
 15 vector<ll>v;
 16 ll n;
 17 ll a[maxn];
 18 ll sum[maxn];
 19 
 20 int tree[maxn<<3];
 21 
 22 int getid(ll x){
 23     return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
 24 }
 25 
 26 void pushup(int rt){
 27     tree[rt]=tree[rt<<1]+tree[rt<<1|1];
 28 }
 29 
 30 void build(int l,int r,int rt){
 31     if(l==r){
 32         tree[rt]=0;
 33         return;
 34     }
 35     int mid=(l+r)/2;
 36     build(lson);
 37     build(rson);
 38     pushup(rt);
 39 }
 40 
 41 void add(int L,int k,int l,int r,int rt){
 42     if(l==r){
 43         tree[rt]+=k;
 44         return;
 45     }
 46     int mid=(l+r)/2;
 47     if(L<=mid) add(L,k,lson);
 48     else add(L,k,rson);
 49     pushup(rt);
 50 }
 51 
 52 ll query(int L,int R,int l,int r,int rt){
 53     if(L<=l&&R>=r){
 54         return tree[rt];
 55     }
 56     int mid=(l+r)/2;
 57     ll ans=0;
 58     if(L<=mid) ans+=query(L,R,lson);
 59     if(R>mid) ans+=query(L,R,rson);
 60     return ans;
 61 }
 62 
 63 
 64 int main(){
 65 
 66     std::ios::sync_with_stdio(false);
 67     ll t;
 68     cin>>n>>t;
 69     for(int i=1;i<=n;i++){
 70         cin>>a[i];
 71     }
 72     v.push_back(t-1);
 73     for(int i=1;i<=n;i++){
 74         sum[i]=a[i]+sum[i-1];
 75         v.push_back(sum[i]);
 76         v.push_back(sum[i]+t-1);
 77     }
 78     if(n==1){
 79         if(a[1]<t){
 80             cout<<1<<endl;
 81         }
 82         else{
 83             cout<<0<<endl;
 84         }
 85     }
 86     else{
 87         sort(v.begin(),v.end());
 88         v.erase(unique(v.begin(),v.end()),v.end());
 89         int Size=v.size();
 90         build(1,Size,1);
 91         add(getid(sum[n]),1,1,Size,1);
 92         ll ans=0;
 93         for(int i=n-1;i>=1;i--){
 94             ans+=query(1,getid(sum[i]+t-1),1,Size,1);
 95             add(getid(sum[i]),1,1,Size,1);
 96         }
 97         ans+=query(1,getid(t-1),1,Size,1);
 98 
 99         cout<<ans<<endl;
100     }
101 
102 }

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转载于:https://www.cnblogs.com/Fighting-sh/p/9719751.html