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Description

有N个位置，M个操作。操作有两种，每次操作如果是1 a b c的形式表示在第a个位置到第b个位置，每个位置加入一个数c 如果是2 a b c形式，表示询问从第a个位置到第b个位置，第C大的数是多少。

Solution

这题是可以区间线段树套权值线段树来做的

但是我想练一下整体二分

顺便写一个树状数组的区间修改+区间查询

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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