HDU 4893(Wow! Such Sequence!-線段樹單點修改+區間求和+改為最近Fib數)

Wow! Such Sequence!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 3830    Accepted Submission(s): 1081

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 F 0 = 1,F 1 = 1,Fibonacci number Fn is defined as F n = F n - 1 + F n - 2 for n ≥ 2.

Nearest Fibonacci number of number x means the smallest Fn where |F n - 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| < 2 31, 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
        

Author Fudan University  

Source 2014 Multi-University Training Contest 3  

Recommend We have carefully selected several similar problems for you:   5426  5425  5424  5423  5422   

存一個sumv,

和改為Fib後的fsum,

mark表示是否改為Fib數

然後線段樹

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEM2(a,i) memset(a,i,sizeof(a));
#define INF (2139062143)
#define MAXN (200000+109)
#define MAXM (100000+10)
typedef __int64 ll;


ll f[100];
const int MFIB = 80;
ll find (ll x) {
	if (x<=1) return 1; 	
	int m=upper_bound(f+1,f+80,x)-(f+1);
	if (llabs(f[m+1]-x)<llabs(f[m]-x)) m++;
	if (llabs(f[m-1]-x)<=llabs(f[m]-x)) m--;
	
	return f[m];
} 



class SegmentTree  
{  
public:  
    ll sumv[MAXN*4],fsum[MAXN*4];
    bool mark[MAXN*4];
    int n;  
    SegmentTree(){}  
    void mem(int _n)  
    {  
    	MEM(sumv) MEM(fsum) MEM(mark)
        n=_n;  
        build(1,1,_n);
    }  

	void build(int o,int L,int R) 
	{
		if (L==R) {
			sumv[o]=mark[o]=0;
			fsum[o]=1; 
		}
		else{
			int M=(R+L)>>1;
			if (L<=M) build(Lson,L,M);
			if (M< R) build(Rson,M+1,R);
		}
		maintain(o,L,R); 
	} 

    void maintain(int o,int L,int R)  
    {
    	if (L<R) 
		{
			sumv[o]=sumv[Lson]+sumv[Rson];
			fsum[o]=fsum[Lson]+fsum[Rson];			
		}		
	}

	int y1,y2,v;
	void update(int o,int L,int R) 
	{
		pushdown(o,L,R);
		if (L==R) {
			sumv[o]+=v;
			fsum[o]=find(sumv[o]);
			mark[o]=0;
			return ;
		}
		else{
			int M=(R+L)>>1;
			if (y1<=M) update(Lson,L,M); 
			if (M< y2) update(Rson,M+1,R);
		}
		maintain(o,L,R); 
		
	}
	void update3(int o,int L,int R) //fib
	{
		pushdown(o,L,R);
		if (y1<=L&&R<=y2) {
			mark[o]=1;
			sumv[o]=fsum[o];
			return ; 
		}
		else{
			int M=(R+L)>>1;
			if (y1<=M) update3(Lson,L,M); //維護pushodown，再次maintain 
			if (M< y2) update3(Rson,M+1,R);
		}
		
		maintain(o,L,R); 
	}
		
	void pushdown(int o,int L,int R) 
	{
		if (L>R) return;
		if (L==R) {
			mark[o]=0;
			return ;
		}   
		if (mark[o])
		{
			mark[Lson]= mark[ Rson ] = 1;
			sumv[Lson]=fsum[Lson]; 
			sumv[Rson]=fsum[Rson]; 
			mark[o]=0;
		}
	}
	ll _sum; 
	
	void query2(int o,int L,int R)
	{
		pushdown(o,L,R);
		if (y1<=L&&R<=y2)
		{
			_sum+=sumv[o];
			return;
		} else {
			int M=(L+R)>>1;
			if (y1<=M) query2(Lson,L,M);
			if (M< y2) query2(Rson,M+1,R);
		}
	}
	

	void add(int l,int r,ll v)
	{
		y1=l,y2=r;this->v=v;
		update(1,1,n);
	}

	void fibset(int l,int r)
	{
		y1=l,y2=r;
		update3(1,1,n);
	}
	
	ll ask(int l,int r)
	{
		_sum=0;
		y1=l,y2=r;
		query2(1,1,n);
		return _sum;
	}
	void print()
	{
		For(i,n)
			cout<<ask(i,i)<<' ';
		cout<<endl;
	
	}

}S;  

int n,m;

int main()
{
//	freopen("data.in","r",stdin);
//	freopen("data.out","w",stdout);

	f[1]=f[2]=1;
	Fork(i,3,80) f[i] = f[i-1] + f[i-2];
	 
//	Fork(i,-10,100) cout<<i<<" "<<find(i)<<endl;
	 
	while(scanf("%d%d",&n,&m)==2) {
		S.mem(n);
		For(i,m) {
			int p;
			scanf("%d",&p);
			if (p==1) {
				int k;ll d;
				scanf("%d%I64d",&k,&d);
				S.add(k,k,d);
			} else {
				int a,b;
				scanf("%d%d",&a,&b);
				if (p==2) printf("%I64d\n",S.ask(a,b));
				else if (p==3) S.fibset(a,b);
			}
//			S.print();
		}
	}
	
	
	
	return 0;
}