Naive Operations

Problem Description

In a galaxy far, far away, there are two integer sequence a and b of length n.
b is a static permutation of 1 to n. Initially a is filled with zeroes.
There are two kind of operations:
1. add l r: add one for al,al+1...ar
2. query l r: query ∑ri=l⌊ai/bi⌋

 

Input

There are multiple test cases, please read till the end of input file.
For each test case, in the first line, two integers n,q, representing the length of a,b and the number of queries.
In the second line, n integers separated by spaces, representing permutation b.
In the following q lines, each line is either in the form 'add l r' or 'query l r', representing an operation.
1≤n,q≤100000, 1≤l≤r≤n, there're no more than 5 test cases.

 

Output

Output the answer for each 'query', each one line.

 

Sample Input

5 12 1 5 2 4 3 add 1 4 query 1 4 add 2 5 query 2 5 add 3 5 query 1 5 add 2 4 query 1 4 add 2 5 query 2 5 add 2 2 query 1 5

 

Sample Output

1 1 2 4 4 6

 

Source

2018 Multi-University Training Contest 2

 

 线段树维护最小值，树状数组维护每点贡献。

我这个线段树写的超级暴力，居然过了....

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#include <iostream>
#include <bits/stdc++.h>
#define maxn 400005
using namespace std;
typedef long long ll;
ll tree[maxn]={0};
ll lazy[maxn]={0};
ll b[100005]={0};
ll c[100005];
int n;
ll lowbit(ll x)
{
    return x&-x;
}
void update(ll x,ll val)
{
    for(int i=x;i<=n;i+=lowbit(i))
        c[i]+=val;
}
ll sum(ll x)
{
    ll res=0;
    for(int i=x;i>0;i-=lowbit(i))
        res+=c[i];
    return res;
}
void pushup(int rt)
{
    tree[rt]=min(tree[rt*2],tree[rt*2+1]);
}
void buildtree(int l,int r,int rt)
{
    if(l==r)
    {
        tree[rt]=b[l];
        return;
    }
    int mid=(l+r)/2;
    buildtree(l,mid,rt*2);
    buildtree(mid+1,r,rt*2+1);
    pushup(rt);
}
void pushdown(int rt)
{
    if(lazy[rt])
    {
        lazy[rt*2]+=lazy[rt];
        lazy[rt*2+1]+=lazy[rt];
        tree[rt*2]+=lazy[rt];
        tree[rt*2+1]+=lazy[rt];
        lazy[rt]=0;
    }
}
void add(int l,int r,int L,int R,int rt,int val)
{
    if(L<=l&&R>=r)
    {
        lazy[rt]+=val;
        tree[rt]+=val;
        return;
    }
    if(lazy[rt]) 
        pushdown(rt);
    int mid=(l+r)/2;
    if(L<=mid) 
        add(l,mid,L,R,rt*2,val);
    if(R>mid) 
    add(mid+1,r,L,R,rt*2+1,val);
    pushup(rt);
}
ll query(int l,int r,int L,int R,int rt)
{
    if(L<=l&&R>=r)
        return tree[rt];
    int mid=(l+r)/2;
    pushdown(rt);
    ll ret=1e9;
    if(L<=mid) 
        ret=query(l,mid,L,R,rt*2);
    if(R>mid) 
    ret=min(ret,query(mid+1,r,L,R,rt*2+1));
    return ret;
}
int get_min(int l,int r,int L,int R,int rt)
{
    if(l==r&&l>=L&&l<=R)
        return l;
    int mid=(l+r)/2;
    int ans;
    if(query(1,n,l,mid,1)<query(1,n,mid+1,r,1)) 
        ans=get_min(l,mid,L,R,rt*2);
    else 
        ans=get_min(mid+1,r,L,R,rt*2+1);
    return ans;
}
int main()
{
    int q;
    while(~scanf("%d%d",&n,&q))
    {
        memset(tree,0,sizeof(tree));
        memset(lazy,0,sizeof(lazy));
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
            scanf("%d",&b[i]);
        buildtree(1,n,1);
        char s[50];
        int l,r;
        while(q--)
        {
           scanf("%s %d %d",s,&l,&r);
           if(s[0]=='a')
           {
               add(1,n,l,r,1,-1);
               while(query(1,n,l,r,1)==0)
               {
                   int id=get_min(1,n,l,r,1);
                   add(1,n,id,id,1,b[id]);
                   update(id,1);
               }
           }
           else
               printf("%lld
",sum(r)-sum(l-1));
       }
    }
    return 0;
}