hdu 4893 Wow! Such Sequence!

Wow! Such Sequence!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1894    Accepted Submission(s): 585

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 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| < 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 12 1 15 41 1 71 3 173 2 42 1 5

 

Sample Output

022

 

题解及代码：

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#define maxn 100110#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;typedef long long ll;ll fibo[100];struct node{    ll s,st,r;} son[maxn<<2];inline void init(){    fibo[0]=1;    fibo[1]=1;    for(int i=2; i<=90; i++)    {        fibo[i]=fibo[i-1]+fibo[i-2];    }}inline ll find_fibo(ll s){    if(s==0) return 1;    int l=0,r=79,mid;    while(l<=r)    {        mid=(l+r)/2;        if(fibo[mid]==s) return fibo[mid];        if(fibo[mid]<s) l=mid+1;        if(fibo[mid]>s) r=mid-1;    }    ll p=s-fibo[l]<0?fibo[l]-s:s-fibo[l];    ll q=s-fibo[r]<0?fibo[r]-s:s-fibo[r];    return p<q?fibo[l]:fibo[r];}inline void PushUp(int rt,int f){    if(f==1||f==3)        son[rt].s=son[rt<<1].s+son[rt<<1|1].s;    if(f==2||f==3)        son[rt].st=son[rt<<1].st+son[rt<<1|1].st;}inline void PushDown(int rt){    if(son[rt].r)    {        son[rt<<1].r=son[rt<<1|1].r=1;        son[rt<<1].s=son[rt<<1].st;        son[rt<<1|1].s=son[rt<<1|1].st;        son[rt].r=0;    }}inline void Build(int l,int r,int rt){    son[rt].r=0;    if(l==r)    {        son[rt].s=0;        son[rt].st=1;        return;    }    int m=(l+r)>>1;    Build(lson);    Build(rson);    PushUp(rt,3);}inline void Update(int p,ll add,int l,int r,int rt){    if(l==r)    {        son[rt].s+=add;        son[rt].st=find_fibo(son[rt].s);        return;    }    PushDown(rt);    int m=(l+r)/2;    if(p<=m) Update(p,add,lson);    else Update(p,add,rson);    PushUp(rt,3);}inline ll Query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)    {        return son[rt].s;    }    PushDown(rt);    int m=(l+r)/2;    ll ret=0;    if(L<=m) ret+=Query(L,R,lson);    if(R>m) ret+=Query(L,R,rson);    PushUp(rt,1);    return ret;}inline void Change(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)    {        son[rt].r=1;        son[rt].s=son[rt].st;        return;    }    PushDown(rt);    int m=(l+r)/2;    if(L<=m) Change(L,R,lson);    if(R>m) Change(L,R,rson);    PushUp(rt,1);}int main(){    init();    int m,n,s,l,r;    while(scanf("%d%d",&n,&m)!=EOF)    {        Build(1,n,1);        for(int i=0; i<m; i++)        {            scanf("%d%d%d",&s,&l,&r);            if(s==1) Update(l,r*1LL,1,n,1);            if(s==2)            {                ll t=Query(l,r,1,n,1);                printf("%I64d\n",t);            }            if(s==3)            {                Change(l,r,1,n,1);            }        }    }    return 0;}/*题解随后补上*/