codeforce1042 D. Petya and Array 树状数组

codeforce1042 D. Petya and Array

相同题型,不过逆序数不用前缀和POJ2299 Ultra-QuickSort 树状数组(求逆序数)+离散化

求区间和小于T的段数

假设i的前缀和为a[i],对于每个a[i]查询get(i)=\sum_{j=0}^{i}(a[j]<a[i]-T),答案为\sum_{i=1}^{n}i-get(i);

准题解:权值线段树(区间第k大)??

#include<bits/stdc++.h>
using namespace std;
const int MAX=2e5+10;
long long hash_[MAX<<1],a[MAX<<1],t;
int c[MAX<<1],n,cnt;
void add(int x,int v)
{
    while(x<=cnt) c[x]+=v,x+=x&(-x);
}
int get(int r)
{
    int sum=0;
    while(r) sum+=c[r],r-=r&(-r);
    return sum;
}
int main()
{
    long long x;
    while(~scanf("%d%lld",&n,&t))
    {
        memset(c,0,sizeof(c));
        memset(a,0,sizeof(a));
        long long ans=0;
        hash_[0]=0;hash_[1]=-t;a[0]=0;
        for(int i=1;i<=n;++i)
        {
            scanf("%lld",&x);
            a[i]+=a[i-1]+x;
            hash_[i*2]=a[i];
            hash_[i*2+1]=a[i]-t;
        }
        sort(hash_,hash_+n*2+2);
        cnt=unique(hash_,hash_+2*n+2)-hash_;
        add(lower_bound(hash_,hash_+cnt,0)-hash_+1,1);
        for(int i=1;i<=n;++i)
        {
            x=lower_bound(hash_,hash_+cnt,a[i]-t)-hash_+1;
            ans+=i-get(x);
            x=lower_bound(hash_,hash_+cnt,a[i])-hash_+1;
            add(x,1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}

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转载自blog.csdn.net/Nrtostp/article/details/82750545
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