# p3863 序列

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define int long long
priority_queue<pair<int,int> >p;
int n,m,bl[100100],a[100100],cnt1,cnt2,b,ans[100100];
int val[100100],w[100100],L[100100],R[100100],tag[100100];
struct upd{
int le,ri,x,t;
};
upd d[100100];
int pl,x,t,id;
};
inline bool cmp(const upd x,const upd y){
if(x.le==y.le)return x.t<y.t;
return x.le<y.le;
}
if(x.pl==y.pl)return x.t<y.t;
return x.pl<y.pl;
}
inline void work(int le,int ri,int x){
int i,j,k;
if(bl[le]==bl[ri]){
for(i=le;i<=ri;i++)val[i]+=x;
k=bl[le];
for(i=L[k];i<=R[k];i++)w[i]=val[i];
sort(w+L[k],w+R[k]+1);
return;
}
for(i=le;i<=R[bl[le]];i++)val[i]+=x;
for(i=L[bl[le]];i<=R[bl[le]];i++)w[i]=val[i];
sort(w+L[bl[le]],w+R[bl[le]]+1);
for(i=L[bl[ri]];i<=ri;i++)val[i]+=x;
for(i=L[bl[ri]];i<=R[bl[ri]];i++)w[i]=val[i];
sort(w+L[bl[ri]],w+R[bl[ri]]+1);
for(i=bl[le]+1;i<bl[ri];i++)tag[i]+=x;
}
inline int go(int wh,int x){
int k=lower_bound(w+L[wh],w+R[wh]+1,x)-w;
if(k==R[wh]+1)return 0;
return R[wh]-k+1;
}
inline int que(int pl,int x){
int i,j,k,res=0;
for(i=1;i<bl[pl];i++)res+=go(i,x-tag[i]);
for(i=L[bl[pl]];i<=pl;i++)if(val[i]>=x-tag[bl[pl]])res++;
return res;
}
signed main(){
int i,j,k;
scanf("%lld%lld",&n,&m);m++;
for(i=1;i<=n;i++)scanf("%lld",&a[i]);
for(i=2;i<=m;i++){
scanf("%lld",&k);
if(k==1){
++cnt2;
scanf("%lld%lld%lld",&d[cnt2].le,&d[cnt2].ri,&d[cnt2].x);
d[cnt2].t=i;
}else {
++cnt1;
scanf("%lld%lld",&q[cnt1].pl,&q[cnt1].x);
q[cnt1].t=i;
q[cnt1].id=cnt1;
}
}
b=sqrt((n+1));j=1;
for(i=1;i<=m;i++)bl[i]=(i-1)/b+1;
int sum=bl[m];
for(i=1;i<=sum;i++)L[i]=R[i-1]+1,R[i]=R[i-1]+b;
R[sum]=min(R[sum],m);
sort(d+1,d+cnt2+1,cmp);
sort(q+1,q+cnt1+1,cmp2);
for(i=1;i<=cnt1;i++){
while(!p.empty()&&-p.top().fi<q[i].pl){
int x=p.top().se;
work(d[x].t,m,-d[x].x);
p.pop();
}
if(q[i].pl!=q[i-1].pl){
work(1,m,a[q[i].pl]);
work(1,m,-a[q[i-1].pl]);
}
while(j<=cnt2&&d[j].le<=q[i].pl){
if(d[j].ri<q[i].pl){j++;continue;}
work(d[j].t,m,d[j].x);
p.push(mp(-d[j].ri,j));
j++;
}
ans[q[i].id]=que(q[i].t-1,q[i].x);
}
for(i=1;i<=cnt1;i++)printf("%lld\n",ans[i]);
return 0;
}

posted @ 2019-10-30 14:23  水题收割者  阅读(159)  评论(0编辑  收藏  举报