树状数组小结

单点修改&区间查询

最普通的树状数组
实现用$lowbit$的优秀性质
原理很简单

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 500005
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define O3 __attribute__((optimize("-O3")))

using namespace std;

ll a[MAXN],tr[MAXN];
ll n,m;

O3 inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
    while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
O3 inline ll lowbit(ll x)
{
    return x&(-x);
}
O3 inline void add(ll x,ll y)
{
    while (x<=n)tr[x]+=y,x+=lowbit(x);
}
O3 inline ll query(ll x)
{
    ll y=0;
    while (x)y+=tr[x],x-=lowbit(x);
    return y;
}
O3 inline ll ask(ll x,ll y)
{
    return query(y)-query(x-1);
}
O3 int main()
{
    //freopen("readin.txt","r",stdin);
    n=read(),m=read();
    fo(i,1,n)a[i]=read(),add(i,a[i]);
    while (m--)
    {
        ll z=read(),x=read(),y=read();
        if (z==1)add(x,y);
        else printf("%lld\n",ask(x,y));
    }
    return 0;
}

---

区间修改&单点查询

实现这个的思路比较巧妙
做一个差分，设$b[i]=a[i]-a[i-1]$，那么${\sum }_{i=1}^{n}b[i]=a[i]$
那么对于一个区间的修改$[x,y]$区间加$k$，相当于$x$位加$k$、$y+1$位减$k$
容易知道这是对的，在$x$前没有影响，从$x$开始才有贡献
用树状数组维护$b$就好了

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 500005
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;++i)
#define O3 __attribute__((optimize("-O3")))

ll a[MAXN],tr[MAXN];
ll n,m;

O3 inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
    while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
O3 inline ll lowbit(ll x)
{
    return x&(-x);
}
O3 inline void change(ll x,ll y)
{
    while (x<=n)tr[x]+=y,x+=lowbit(x);
}
O3 inline ll query(ll x)
{
    ll y=0;
    while (x)y+=tr[x],x-=lowbit(x);
    return y;
}
O3 int main()
{
    //freopen("readin.txt","r",stdin);
    n=read(),m=read();
    fo(i,1,n)a[i]=read(),change(i,a[i]-a[i-1]);
    while (m--)
    {
        ll z=read(),x,y;
        if (z==1)
        {
            x=read(),y=read(),z=read();
            change(x,z),change(y+1,-z);
        }
        else printf("%lld\n",query(read()));
    }
    return 0;
}

---

伪线段树树状数组

其实这个就是支持区间修改和区间查询的树状数组
对于一段区间查询$[x,y]$，拆成$[1,x-1]$和$[1,y]$，单独考虑$[1,x]$的答案即可

• 那么$ans={\sum }_{i=1}^{x}a[i]=a[1]+a[2]+a[3]+...+a[x]$

• $=(b[1])+(b[1]+b[2])+(b[1]+b[2]+b[3])+...+(b[1]+b[2]+b[3]...+b[x])$

• $=b[1]\ast x+b[2]\ast (x-1)+b[3]\ast (x-2)+...+b[x]$

• $=x(b[1]+b[2]+b[3]+...+b[x])-(b[1]\ast 0+b[2]\ast 1+b[3]\ast 2+...+b[x]\ast (x-1))$

再设一个$c$数组，使$c[i]=b[i]\ast (i-1)$
那么$[1,x]$的和即为$x\ast query(b,x)-query(c,x)$

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 500005
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;++i)
#define O3 __attribute__((optimize("-O3")))

ll a[MAXN],tr[MAXN],tr1[MAXN];
ll n,m;

O3 inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
    while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
O3 inline ll lowbit(ll x)
{
    return x&(-x);
}
O3 inline void change(ll *tr,ll x,ll y)
{
    while (x<=n)tr[x]+=y,x+=lowbit(x);
}
O3 inline ll query(ll *tr,ll x)
{
    ll y=0;
    while (x)y+=tr[x],x-=lowbit(x);
    return y;
}
O3 int main()
{
    //freopen("readin.txt","r",stdin);
    n=read(),m=read();
    fo(i,1,n)a[i]=read(),change(tr,i,a[i]-a[i-1]),change(tr1,i,(a[i]-a[i-1])*(i-1));
    while (m--)
    {
        ll z=read(),x,y;
        if (z==1)
        {
            x=read(),y=read(),z=read();
            change(tr,x,z),change(tr,y+1,-z);
            change(tr1,x,z*(x-1)),change(tr1,y+1,-z*y);
        }
        else x=read(),y=read(),
        printf("%lld\n",(y*query(tr,y)-(x-1)*query(tr,x-1))-(query(tr1,y)-query(tr1,x-1)));
    }
    return 0;
}