几个树状数组的简单题

#include<iostream>
#include<stdio.h>
#include<vector>
#include<queue>
#include<algorithm>
#include<memory.h>
#include<string.h>
#include<cmath>
#include<map>
#include<set>
#define INF 0x3f3f3f3f
#define PII pair<int,int>
#define MOD 1000000000
#define MAX_N 1000
using namespace std;
int n,m,w,x,y,tmp;
int xx[100005];
int a[100005];
int ans[100005];
int max_n=0;
int lowbit(int x)
{
    return x&(-x);
}

int sum(int x)
{
    int res=0;
    while(x)
    {
        res+=a[x];
        x-=lowbit(x);
    }
    return res;
}

void add(int x)
{
    while(x<=max_n)
    {
        a[x]++;
        x+=lowbit(x);
    }
}

int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>x>>y;
        xx[i]=x+1;
        max_n=max(x+1,max_n);
    }

    for(int i=1;i<=n;i++)
    {
        int t=sum(xx[i]);
        add(xx[i]);
        ans[t]++;
    }

    for(int i=0;i<=n-1;i++)
        cout<<ans[i]<<endl;
    return 0;
}

#include<iostream>
#include<stdio.h>
#include<vector>
#include<queue>
#include<algorithm>
#include<memory.h>
#include<string.h>
#include<cmath>
#include<map>
#include<set>
#define INF 0x3f3f3f3f
#define PII pair<int,int>
#define MOD 1000000000
#define MAX_N 1000
using namespace std;
int n,m,w,x,y,tmp;
int xx[100005];
int a[100005];
int ans[100005];
int max_n=0;
int lowbit(int x)
{
    return x&(-x);
}

int sum(int x)
{
    int res=0;
    while(x)
    {
        res+=a[x];
        x-=lowbit(x);
    }
    return res;
}

void add(int x,int num)
{
    while(x<=n)
    {
        a[x]+=num;
        x+=lowbit(x);
    }
}

int main()
{
    //cout<<6+lowbit(6)<<endl;
    memset(a,0,sizeof(a));
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>x>>y;
        //xx[i]=x;
        add(x,1);
        add(y+1,-1);
        //max_n=max(x,max_n);
    }
    for(int i=1;i<=n;i++)
        if(i!=n)
            cout<<sum(i)<<" ";
        else
            cout<<sum(i)<<endl;

    //cout<<res<<endl;
    return 0;
}

#include<iostream>
#include<algorithm>
#include<memory.h>
using namespace std;
long long a[500005];
long long b[500005];
long long c[500005];
long long n,t;
long long lowbit(long long x)
{
    return x&(-x);
}
long long sum(long long x)
{
    long long res=0;
    while(x)
    {
        res+=c[x];
        x-=lowbit(x);
    }
    return res;
}
void add(long long x)
{
    while(x<=n)
    {
        c[x]++;
        x+=lowbit(x);
    }
}

int main()
{
    memset(c,0,sizeof(c));
    cin>>n;
    for(long long i=1;i<=n;i++)
    {
        cin>>a[i];
        b[i]=a[i];
    }
    sort(a+1,a+1+n);
    long long res=0;
    for(long long i=1;i<=n;i++)
    {
        //index = 小于b[i]的元素个数+1
        //即b[i] 被映射成了 index
        long long index=lower_bound(a+1,a+1+n,b[i])-a;
        long long upper_num=i-1-sum(index-1);
        add(index);
        res+=upper_num;
    }
    cout<<res<<endl;
    return 0;
}

#include<iostream>
#include<stdio.h>
#include<memory.h>
using namespace std;
int a[500005];
int c[500005];
int flag[500005];
int res[500005];
int n,t;
int lowbit(int x)
{
    return x&(-x);
}

int sum(int x)
{
    int res=0;
    while(x)
    {
        res+=c[x];
        x-=lowbit(x);
    }
    return res;
}

void add(int x,int num)
{
    while(x<=n)
    {
        c[x]+=num;
        x+=lowbit(x);
    }
}
int main()
{
    memset(c,0,sizeof(c));
    memset(flag,0,sizeof(flag));
    scanf("%d",&n);
    for(int i=2;i<=n;i++)
        scanf("%d",&a[i]);

    //最后一个点的编号
    res[n]=a[n]+1;
    add(res[n],1);
    flag[res[n]]=1;

    for(int i=n-1;i>1;i--)
    {
        //初始位置
        int t=a[i]+1;
        int l=0,r=n,pianyi;
        //二分枚举偏移量
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(mid+t>n)
            {
                mid=n-t;
                r=2*(n-t)-l;
            }
            int k=sum(mid+t);
            if(mid>k)
            {
                r=mid-1;
            }
            if(mid<k)
            {
                l=mid+1;
            }
            if(mid==k)
            {
                pianyi=mid;
                r=mid-1;
            }
        }
        res[i]=t+pianyi;
        flag[res[i]]=1;
        add(res[i],1);
    }

    for(int i=1;i<=n;i++)
    {
        if(!flag[i])
        {
            res[1]=i;
            break;
        }
    }
    for(int i=1;i<=n;i++)
        printf("%d\n",res[i]);
    return 0;
}