HDU6315:Naive Operations 题解

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⌋ ∑ i = l r ⌊ a i / b i ⌋

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

---

考虑到b是一个1~n的排列，1这个数最多会贡献n，2这个数最多会贡献n/2，所以根据调和级数，总贡献是nlogn级别的
我们考虑建线段树，线段树的每个节点初始值是b数组中的值，每次add操作相当于一个区间减操作，如果当前区间的最小值大于1的话就直接更新懒标记，否则递归下去，直到那个是1的点，贡献+1，值置为b[i]，因为至多有nlogn次暴力访问到叶子的操作，所以总复杂度 O(nlog2n) O ( n l o g 2 n )

#include 
using namespace std;

#define LL long long
#define LB long double
#define ull unsigned long long
#define x first
#define y second
#define pb push_back
#define pf push_front
#define mp make_pair
#define Pair pair<int,int>
#define pLL pair
#define pii pair

const int INF=2e9;
const LL LINF=2e16;
const int magic=348;
const int MOD=998244353;
const double eps=1e-10;
const double pi=acos(-1);

inline int getint()
{
    bool f;char ch;int res;
    while (!isdigit(ch=getchar()) && ch!='-') {}
    if (ch=='-') f=false,res=0; else f=true,res=ch-'0';
    while (isdigit(ch=getchar())) res=res*10+ch-'0';
    return f?res:-res;
}

const int MAXN=1e5;

int n,q;
int b[MAXN+48];

namespace SegmentTree
{
    struct node
    {
        int left,right;
        int sum,lazy,minn;
    }tree[MAXN*4];
    inline void pushup(int cur)
    {
        tree[cur].minn=min(tree[cur<<1].minn,tree[cur<<1|1].minn);
        tree[cur].sum=tree[cur<<1].sum+tree[cur<<1|1].sum;
    }
    inline void pushdown(int cur)
    {
        tree[cur<<1].lazy+=tree[cur].lazy;
        tree[cur<<1|1].lazy+=tree[cur].lazy;
        tree[cur<<1].minn-=tree[cur].lazy;
        tree[cur<<1|1].minn-=tree[cur].lazy;
        tree[cur].lazy=0;
    }
    inline void build(int cur,int left,int right)
    {
        tree[cur].left=left;tree[cur].right=right;
        tree[cur].lazy=0;
        if (left!=right)
        {
            int mid=(left+right)>>1;
            build(cur<<1,left,mid);
            build(cur<<1|1,mid+1,right);
            pushup(cur);
        }
        else
        {
            tree[cur].sum=0;tree[cur].minn=b[left];
        }
    }
    inline void update(int cur,int left,int right)
    {
        if (tree[cur].left==tree[cur].right)
        {
            if (tree[cur].minn==1)
            {
                tree[cur].sum++;
                tree[cur].minn=b[tree[cur].left];
                return;
            }
            tree[cur].minn--;
            return;
        }
        if (left<=tree[cur].left && tree[cur].right<=right && tree[cur].minn>1)
        {
            tree[cur].lazy++;tree[cur].minn--;
            return;
        }
        pushdown(cur);
        int mid=(tree[cur].left+tree[cur].right)>>1;
        if (left<=mid) update(cur<<1,left,right);
        if (mid+1<=right) update(cur<<1|1,left,right);
        pushup(cur);
    }
    inline int query(int cur,int left,int right)
    {
        if (left>tree[cur].right || rightleft || left>right) return 0;
        if (left<=tree[cur].left && tree[cur].right<=right) return tree[cur].sum;
        pushdown(cur);int res=query(cur<<1,left,right)+query(cur<<1|1,left,right);
        pushup(cur);return res;
    }
}

int main ()
{
    int i,l,r;char op[10];
    while (scanf("%d%d",&n,&q)!=EOF)
    {
        for (i=1;i<=n;i++) b[i]=getint();
        SegmentTree::build(1,1,n);
        while (q--)
        {
            scanf("%s%d%d",op+1,&l,&r);
            if (op[1]=='a') SegmentTree::update(1,l,r); else printf("%d\n",SegmentTree::query(1,l,r));
        }
    }
    return 0;
}