HDU-6315 Naive Operations 線段樹

Naive Operations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)
Total Submission(s): 444    Accepted Submission(s): 138

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

題意:給出一個長度為n初值為0的數組，以及長度為n的b數組，然后q次操作，add(l,r) 使得區間l~r所有元素+1，或者查詢l~r區間a[i]/b[i]的和

思路：維護區間的a的最大值和b的最小值，使用lazy標記就不需要更新到每個點了。每次更新當當前最大值a小於最小的b，那么下面子節點就都不需要查詢了

大於時我們暴力找到l==r然后cnt+1,同時分母+上b[l]，比如當前是2/2 我們就把它變成 2/4 這樣我們還需要增加2次才能到達4/4 再變成4/6.

#include<bits/stdc++.h>
using namespace std;
#define Lson l,m,rt<<1
#define Rson m+1,r,rt<<1|1
#define pll pair<int,int>
#define mp make_pair
#define ll long long
#define INF 0x3f3f3f3f
const int maxn=1e5+5;
int n,q;
struct node
{
    int cnt,addv,minb,maxa;
} tree[maxn<<2];
int b[maxn];
void push_up(int rt)
{
    tree[rt].minb=min(tree[rt<<1].minb,tree[rt<<1|1].minb);
    tree[rt].cnt=tree[rt<<1].cnt+tree[rt<<1|1].cnt;
    tree[rt].maxa=max(tree[rt<<1].maxa,tree[rt<<1|1].maxa);
}
void push_down(int rt)
{
    if(tree[rt].addv)
    {
        int v=tree[rt].addv;
        tree[rt].addv=0;
        tree[rt<<1].maxa+=v;
        tree[rt<<1|1].maxa+=v;
        tree[rt<<1].addv+=v;
        tree[rt<<1|1].addv+=v;
    }
}
void build(int l,int r,int rt)
{
    tree[rt].addv=0;
    if(l==r)
    {
        tree[rt].cnt=tree[rt].maxa=0;
        tree[rt].minb=b[l];
        return;
    }
    int m=l+r>>1;
    build(Lson);
    build(Rson);
    push_up(rt);
}
void updata(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        tree[rt].maxa++;
        if(tree[rt].maxa<tree[rt].minb)
        {
            tree[rt].addv++;
            return;
        }
        if(l==r&&tree[rt].maxa>=tree[rt].minb)
        {
            tree[rt].cnt++;
            tree[rt].minb+=b[l];
            return;
        }
    }
    push_down(rt);
    int m=l+r>>1;
    if(L<=m)
        updata(L,R,Lson);
    if(R>m)
        updata(L,R,Rson);
    push_up(rt);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        return tree[rt].cnt;
    }
    int m=l+r>>1;
    push_down(rt);
    int ans=0;
    if(L<=m)
        ans+=query(L,R,Lson);
    if(R>m)
        ans+=query(L,R,Rson);
    return ans;
}
int main()
{
    int n,x,y;
    while(~scanf("%d %d",&n,&q))
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&b[i]);
        }
        build(1,n,1);
        char pp[6];
        int l,r;
        while(q--)
        {
            scanf("%s%d%d",pp,&l,&r);
            if(pp[0]=='a')
            {
                updata(l,r,1,n,1);
            }
            else
            {
                printf("%d\n",query(l,r,1,n,1));
            }
        }
    }
    return 0;
}

 

 

 

 

PS：摸魚怪的博客分享，歡迎感謝各路大牛的指點~