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Description

有N個位置，M個操作。操作有兩種，每次操作如果是1 a b c的形式表示在第a個位置到第b個位置，每個位置加入一個數c 如果是2 a b c形式，表示詢問從第a個位置到第b個位置，第C大的數是多少。

Solution

這題是可以區間線段樹套權值線段樹來做的

但是我想練一下整體二分

順便寫一個樹狀陣列的區間修改+區間查詢

class BIT
{
    #define NM 50005
    #define lb(x) (x&(-x))
    private:
        ll t1[NM],t2[NM],N;
        BIT(){}
    public:
        BIT(int _n):N(_n){memset(t1,0,sizeof t1);memset(t2,0,sizeof t2);}
        inline void CC(int p,int v){for(reg int x=p;x<=N;x+=lb(x))t1[x]+=v,t2[x]+=v*p*1ll;}
        inline void C(int l,int r,int x){CC(l,x);CC(r+1,-x);}
        inline ll GG(int p){ll r=0;for(reg int x=p;x;x-=lb(x))r+=(p+1)*t1[x]-t2[x];return r;}
        inline ll G(int l,int r){return GG(r)-GG(l-1);}
    #undef NM
    #undef lb
};
 

Code 

#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}
#define reg register
class BIT
{
    #define NM 50005
    #define lb(x) (x&(-x))
    private:
        ll t1[NM],t2[NM],N;
        BIT(){}
    public:
        BIT(int _n):N(_n){memset(t1,0,sizeof t1);memset(t2,0,sizeof t2);}
        inline void CC(int p,int v){for(reg int x=p;x<=N;x+=lb(x))t1[x]+=v,t2[x]+=v*p*1ll;}
        inline void C(int l,int r,int x){CC(l,x);CC(r+1,-x);}
        inline ll GG(int p){ll r=0;for(reg int x=p;x;x-=lb(x))r+=(p+1)*t1[x]-t2[x];return r;}
        inline ll G(int l,int r){return GG(r)-GG(l-1);}
    #undef NM
    #undef lb
};
#define MN 50005
struct ques{int l,r,id,opt;ll c;}q[MN],b1[MN],b2[MN];
int n,m,tot,cnt,num[MN],Ans[MN];
void solve(int l=1,int r=tot,int ql=1,int qr=m)
{
    if(ql>qr) return;
//  printf("%d %d %d %d\n",l,r,ql,qr);
    static BIT T(n);register int i;
    if(l==r)
    {
        for(i=ql;i<=qr;++i)if(q[i].opt==2)Ans[q[i].id]=num[l];
        return;
    }
    register int mid=(l+r+1)>>1,tpb1=0,tpb2=0;register ll tmp;
    for(i=ql;i<=qr;++i)
    {
        if(q[i].opt==1)
        {
            if(q[i].c>=num[mid]) T.C(q[i].l,q[i].r,1),b2[++tpb2]=q[i];
            else b1[++tpb1]=q[i];
        }
        else
        {
            tmp=T.G(q[i].l,q[i].r);
            if(tmp<q[i].c) q[i].c-=tmp,b1[++tpb1]=q[i];
            else b2[++tpb2]=q[i];
        }
    }
    for(i=ql;i<=qr;++i)if(q[i].c>=num[mid]&&q[i].opt==1) T.C(q[i].l,q[i].r,-1);
    bool has1=false,has2=false;
    for(i=1;i<=tpb1;++i) q[i+ql-1]=b1[i];
    for(i=1;i<=tpb2;++i) q[qr-tpb2+i]=b2[i];
    solve(l,mid-1,ql,ql+tpb1-1);solve(mid,r,qr-tpb2+1,qr);
}
int main()
{
//  freopen("testdata.in","r",stdin);
//  freopen("testdata.out","w",stdout);
    n=read();m=read();
    register int i;
    for(i=1;i<=m;++i)
    {
        q[i].opt=read(),q[i].l=read(),q[i].r=read(),q[i].c=read();
        if(q[i].opt==1) num[++tot]=q[i].c;
        if(q[i].opt==2) q[i].id=++cnt;
    }
    std::sort(num+1,num+tot+1);
    tot=std::unique(num+1,num+tot+1)-num-1;
    solve();
    for(i=1;i<=cnt;++i) printf("%d\n",Ans[i]);
    return 0;
}
 

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