首页 > 代码库 > HDU4893:Wow! Such Sequence!(线段树lazy)

HDU4893:Wow! Such Sequence!(线段树lazy)

2024-07-14 21:18:13   227人阅读

Problem Description
Recently, Doge got a funny birthday present from his new friend, Protein Tiger from St. Beeze College. No, not cactuses. It‘s a mysterious blackbox.

After some research, Doge found that the box is maintaining a sequence an of n numbers internally, initially all numbers are zero, and there are THREE "operations":

1.Add d to the k-th number of the sequence.
2.Query the sum of ai where l ≤ i ≤ r.
3.Change ai to the nearest Fibonacci number, where l ≤ i ≤ r.
4.Play sound "Chee-rio!", a bit useless.

Let F0 = 1,F1 = 1,Fibonacci number Fn is defined as Fn = Fn - 1 + Fn - 2 for n ≥ 2.

Nearest Fibonacci number of number x means the smallest Fn where |Fn - x| is also smallest.

Doge doesn‘t believe the machine could respond each request in less than 10ms. Help Doge figure out the reason.
 

Input
Input contains several test cases, please process till EOF.
For each test case, there will be one line containing two integers n, m.
Next m lines, each line indicates a query:

1 k d - "add"
2 l r - "query sum"
3 l r - "change to nearest Fibonacci"

1 ≤ n ≤ 100000, 1 ≤ m ≤ 100000, |d| < 231, all queries will be valid.
 

Output
For each Type 2 ("query sum") operation, output one line containing an integer represent the answer of this query.
 

Sample Input
1 1
2 1 1
5 4
1 1 7
1 3 17
3 2 4
2 1 5
 

Sample Output
0
22


#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define ll __int64
#define maxn 100005
#define ls l,mid,2*i
#define rs mid+1,r,2*i+1
#define lson 2*i
#define rson 2*i+1
struct node
{
    int l,r;
    ll e,f;//e为该区间的和，f为其最近的斐波那契数
    int flag,len;//flag,标记这个区间内是否斐波那契数，len为长度
} a[maxn<<2];
int n,m;
ll f[90] = {1,1};

ll pabs(ll a)
{
    return a<0?-a:a;
}

void PushDown(int i,int m)
{
    if(a[i].flag)
    {
        a[lson].flag = a[rson].flag = a[i].flag;
        a[lson].len = a[i].flag*(m-m>>1);
        a[rson].len = a[i].flag*(m>>1);
        a[lson].e = a[lson].f;
        a[rson].e = a[rson].f;
        a[i].flag = 0;
    }
}

void PushUp(int i)
{
    a[i].e = a[lson].e+a[rson].e;
    a[i].f = a[lson].f+a[rson].f;
}

void init(int l,int r,int i)
{
    a[i].flag = a[i].len = 0;
    a[i].l = l;
    a[i].r = r;
    a[i].e = 0;
    if(l == r)
    {
        a[i].f = 1;
        return;
    }
    int mid = (l+r)>>1;
    init(ls);
    init(rs);
    PushUp(i);
}

void add(int pos,int m,int l,int r,int i)
{
    if(pos<l || pos>r) return ;
    if(l == r)
    {
        if(a[i].flag)
        {
            a[i].e = m+a[i].f;
            a[i].flag = 0;
            a[i].len = 0;
        }
        else a[i].e+=m;
        int p = lower_bound(f,f+80,a[i].e)-f;
        if(!p)
            a[i].f = 1;
        else if(pabs(a[i].e-f[p])<pabs(a[i].e-f[p-1]))
            a[i].f = f[p];
        else
            a[i].f = f[p-1];
        return ;
    }
    PushDown(i,r-l+1);
    int mid = (l+r)>>1;
    if(pos<=mid) add(pos,m,ls);
    else add(pos,m,rs);
    PushUp(i);
}

ll query(int L,int R,int l,int r,int i)
{
    if(R<l || L>r) return 0;
    else if(L<=l && R>=r) return a[i].e;
    PushDown(i,r-l+1);
    ll ans = 0;
    int mid = (l+r)>>1;
    if(L<=mid)
        ans += query(L,R,ls);
    if(R>mid)
        ans += query(L,R,rs);
    return ans;
}

void change(int L,int R,int l,int r,int i)
{
    if(R<l || L>r) return ;
    if(L<=l && R>=r)
    {
        a[i].e = a[i].f;
        a[i].flag = 1;
        a[i].len = r-l+1;
        return ;
    }
    PushDown(i,r-l+1);
    int mid = (l+r)>>1;
    if(L<=mid)
        change(L,R,ls);
    if(R>mid)
        change(L,R,rs);
    PushUp(i);
}

int main()
{
    int i,j,x,k,d,l,r;
    for(i = 2; i<80; i++)
        f[i] = f[i-1]+f[i-2];
    while(~scanf("%d%d",&n,&m))
    {
        init(1,n,1);
        while(m--)
        {
            scanf("%d",&x);
            if(x == 1)
            {
                scanf("%d%d",&k,&d);
                add(k,d,1,n,1);
            }
            else
            {
                scanf("%d%d",&l,&r);
                if(x == 2)
                    printf("%I64d\n",query(l,r,1,n,1));
                else
                    change(l,r,1,n,1);
            }
        }
    }

    return 0;
}




联系
我们