P2068 statistics and P2068 statistics

P2068 statistics and P2068 statistics
Description

Given a sequence with a length of n (n <= 100000) and an initial value of 0, x (x <= 10000) is used to modify numbers at certain positions, and a number is added each time, then, the y (y <= 10000) problem is raised, and the sum of each section is obtained. The time limit is 1 second.

Input/Output Format

Input Format:

 

Number of 1 in the first row, indicating the length of the sequence n

Number of 1 operations in the second row, w

Rows w are followed in sequence, indicating join and query operations respectively.

Here, x is added, and y is used to represent the query.

The format of x is "x a B", indicating that B is added to the position of sequence.

The format of y is "y a B", which indicates the sum of a and B.

 

Output Format:

 

Each row contains one number, which is the result of each query.

 

Input and Output sample input sample #1:

54x 3 8y 1 3x 4 9y 3 4

Output sample #1:

817

Bare line segment tree!

  1 #include<iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<cmath>  5 #include<queue>  6 #include<algorithm>  7 #define ls k<<1  8 #define rs k<<1|1  9 using namespace std; 10 const int MAXN=100001; 11 void read(int &n) 12 { 13     char c='+';int x=0;bool flag=0; 14     while(c<'0'||c>'9') 15     {c=getchar();if(c=='-')flag=1;} 16     while(c>='0'&&c<='9') 17     {x=x*10+c-48,c=getchar();} 18     flag==1?n=-x:n=x; 19 } 20 int n,m,chair; 21 int ans=-1; 22 int now=0; 23 int you,mei; 24 struct node 25 { 26     int l,r,w,fm; 27 }tree[MAXN*4]; 28 void update(int k) 29 { 30     tree[k].w=tree[ls].w+tree[rs].w; 31 } 32 void pushdown(int k) 33 { 34     tree[ls].w+=(tree[ls].r-tree[ls].l+1)*tree[k].fm; 35     tree[rs].w+=(tree[rs].r-tree[rs].l+1)*tree[k].fm; 36     tree[ls].fm+=tree[k].fm; 37     tree[rs].fm+=tree[k].fm; 38     tree[k].fm=0; 39     //cout<<"ls:"<<(ls)<<" "<<tree[ls].w<<" "; 40     //cout<<"rs:"<<(rs)<<" "<<tree[rs].w<<" "<<endl; 41 } 42 void build_tree(int ll,int rr,int k) 43 { 44     tree[k].l=ll;tree[k].r=rr; 45     if(ll==rr) 46     { 47         tree[k].w=0; 48         return ; 49     } 50     int mid=(ll+rr)>>1; 51     build_tree(ll,mid,ls); 52     build_tree(mid+1,rr,rs); 53     update(k); 54 } 55 void interval_ask(int ll,int rr,int k) 56 { 57     if(ll>tree[k].r||rr<tree[k].l) 58         return ; 59     if(ll<=tree[k].l&&tree[k].r<=rr) 60     { 61         ans+=tree[k].w; 62         return ; 63     } 64     int mid=(tree[k].l+tree[k].r)>>1; 65     if(tree[k].fm) 66     pushdown(k); 67     if(ll<=mid) 68     interval_ask(ll,rr,ls); 69     if(rr>mid) 70     interval_ask(ll,rr,rs); 71     if(tree[k].fm) 72     pushdown(k); 73 } 74 void point_change(int pos,int v,int k) 75 { 76     if(pos>tree[k].r||pos<tree[k].l) 77         return ; 78     if(tree[k].l==tree[k].r) 79     { 80         tree[k].w+=v; 81         return ; 82     } 83     point_change(pos,v,ls); 84     point_change(pos,v,rs); 85     update(k); 86 } 87 void interval_change(int ll,int rr,int num,int k) 88 { 89     if(ll>tree[k].r||rr<tree[k].l) 90         return ; 91     if(ll<=tree[k].l&&rr>=tree[k].r) 92     { 93         tree[k].w+=num; 94         tree[k].fm+=(tree[k].r-tree[k].l+1)*num; 95         return ; 96     } 97     int mid=(tree[k].l+tree[k].r)>>1; 98     if(tree[k].fm) 99     pushdown(k);100     if(ll<=mid)101     interval_change(ll,rr,num,ls);102     if(rr>mid)103     interval_change(ll,rr,num,rs);104     update(k);105 }106 int main()107 {108     read(n);read(m);109     build_tree(1,n,1);110     for(int i=1;i<=m;i++)111     {112         char how;113         cin>>how;114         if(how=='x')115         {116             int pos,v;117             read(pos);read(v);118             point_change(pos,v,1);119         }120         else 121         {122             ans=0;123             int x,y;124             read(x);read(y);125             interval_ask(x,y,1);126             printf("%d\n",ans);127         }128     }129     return 0;130 }