Luogu P2309 loidc,賣賣萌

題目連接：Click here

題目大意：給你一個長度爲n的數串，問這個數串的sum爲正數的子串個數

Solution:

咱們先處理如下前綴和，記爲\(s_i\)

則問題能夠轉化爲求有多少對\(i,j\)知足\(j>i,s_j-s_{i-1}>0\)

咱們把全部的\(s_i\)變成\(-s_i\)，則問題就變成了知足\(i<j,s_{i-1}>s_j\)的點對個數

而後用樹狀數組求逆序對便可

Code:

#include<bits/stdc++.h>
#define lowbit(x) x&(-x)
using namespace std;
const int N=1e5+2;
long long ans;
int n,a[N],tree[N];
struct Pos{int id,val;}p[N];
inline bool cmp(Pos a,Pos b){return a.val==b.val?a.id<b.id:a.val<b.val;}
void add(int x){for(int i=x;i<=n+1;i+=lowbit(i))tree[i]++;}
int sum(int x){int re=0;for(int i=x;i>=1;i-=lowbit(i)) re+=tree[i];return re;}
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return x*f;
}
int main(){
    n=read();
    for(int i=1;i<=n;i++){
        a[i]=read();
        p[i].val=p[i-1].val-a[i];
        p[i].id=i+1;
    }p[0].id=1;sort(p,p+n+1,cmp);
    for(int i=0;i<=n;i++){
        ans+=sum(n+1)-sum(p[i].id);
        add(p[i].id);
    }printf("%lld\n",ans);
    return 0;
}