[Noi38] game

Maintenance information within a period of two sections with a number of segments: 1. How much experience you need to have a person can make upgrades, 2. grade and. A single point can be modified to do direct violence, interval modify this section will not be considered if the upgrade is not marked but recursion lazy mark.

Consider the time complexity of: a single point modification time complexity is $ O (LOG_ {2} n-) $ , and the interval modification since each point up to upgrade n + q , with the total time complexity is $ O (( + Q n-) LOG_ m {2}) $ , you can lead.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define N 1000001
 4 #define L (k<<1)
 5 #define R (L+1)
 6 #define mid (l+r>>1)
 7 int n,m,q,p,x,y,z,mi[N],f[N],lazy[N],a[N];
 8 void up(int k){
 9     f[k]=f[L]+f[R];
10     mi[k]=min(mi[L],mi[R]);
11 }
12 void down(int k){
13     lazy[L]+=lazy[k];
14     mi[L]-=lazy[k];
15     lazy[R]+=lazy[k];
16     mi[R]-=lazy[k];
17     lazy[k]=0;
18 }
19 void update(int k,int l,int r,int x,int y,int z){
20     if ((l>y)||(x>r))return;
21     if ((x<=l)&&(r<=y)){
22         mi[k]-=z;
23         if (mi[k]>0){
24             lazy[k]+=z;
25             return;
26         }
27         if (l==r){
28            mi[k]-=a[f[k]+1];
29            f[k]=upper_bound(a+1,a+m+1,-mi[k])-a-1;
30            mi[k]+=a[f[k]+1];
31             return;
32         }
33     }
34     down(k);
35     update(L,l,mid,x,y,z);
36     update(R,mid+1,r,x,y,z);
37     up(k);
38 }
39 void update(int k,int l,int r,int x,int y){
40     if (l==r){
41         f[k]=upper_bound(a+1,a+m+1,y)-a-1;
42         mi[k]=a[f[k]+1]-y;
43         return;
44     }
45     down(k);
46     if (x<=mid)update(L,l,mid,x,y);
47     else update(R,mid+1,r,x,y);
48     up(k);
49 }
50 int query(int k,int l,int r,int x,int y){
51     if ((l>y)||(x>r))return 0;
52     if ((x<=l)&&(r<=y))return f[k];
53     down(k);
54     return query(L,l,mid,x,y)+query(R,mid+1,r,x,y);
55 }
56 int main(){
57     scanf("%d%d%d",&n,&m,&q);
58     for(int i=1;i<=m;i++)scanf("%d",&a[i]);
59     a[m+1]=0x3f3f3f3f;
60     for(int i=1;i<=n;i++){
61         scanf("%d",&x);
62         update(1,1,n,i,x);
63     }
64     for(int i=1;i<=q;i++){
65         scanf("%d%d%d",&p,&x,&y);
66         if (p==1){
67             scanf("%d",&z);
68             update(1,1,n,x,y,z);
69         }
70         if (p==2)update(1,1,n,x,y);
71         if (p==3)printf("%d\n",query(1,1,n,x,y));
72     }
73 }

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