HDU-6315 Naive Operations 线段树

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

题意:给出一个长度为n初值为0的数组，以及长度为n的b数组，然后q次操作，add(l,r) 使得区间l~r所有元素+1，或者查询l~r区间a[i]/b[i]的和

思路：维护区间的a的最大值和b的最小值，使用lazy标记就不需要更新到每个点了。每次更新当当前最大值a小于最小的b，那么下面子节点就都不需要查询了

大于时我们暴力找到l==r然后cnt+1,同时分母+上b[l]，比如当前是2/2 我们就把它变成 2/4 这样我们还需要增加2次才能到达4/4 再变成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
",query(l,r,1,n,1));
            }
        }
    }
    return 0;
}

PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~