hdu 4893 线段树

Wow! Such Sequence!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 620    Accepted Submission(s): 191

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

 

Source

2014 Multi-University Training Contest 3

#include<stdio.h>#define N 100100#include<math.h>typedef __int64 ll;#define inf ((ll)1<<63)-1struct node{    ll x,y,p,sum;}a[N*3];ll f[100];void fun(){    ll i;    f[0]=1; f[1]=1;    for(i=2;i<=91;i++)    {        f[i]=f[i-1]+f[i-2];       //printf("%I64d\n",f[i]);    }}ll binary(ll l,ll r,ll k){    if(k<0)  return 1;    ll i,aa,ans=inf,u;    //printf("%I64d\n",ans);    for(i=0;i<=91;i++)    {        aa=f[i]-k;        if(aa<0)            aa=-aa;        if(aa<ans)        {            ans=aa;            u=f[i];        }    }    return u;}void build(ll t,ll x,ll y){    a[t].x=x; a[t].y=y; a[t].sum=0;  a[t].p=0;    if(x==y)  return;    ll mid=(a[t].x+a[t].y)>>1,temp=t<<1;    build(temp,x,mid);    build(temp+1,mid+1,y);}void update(ll t,ll k,ll d){    if(a[t].x==a[t].y)    {        a[t].sum+=d;        a[t].p=0;        return;    }    ll temp=t<<1,mid=(a[t].x+a[t].y)>>1;    if(k<=mid)        update(temp,k,d);    else        update(temp+1,k,d);    a[t].sum=a[temp].sum+a[temp+1].sum;    a[t].p=(a[temp].p)&(a[temp+1].p);}ll query(ll t ,ll x,ll y){    if(a[t].x==x&&a[t].y==y)        return a[t].sum;    ll mid=(a[t].x+a[t].y)>>1,temp=t<<1;    if(y<=mid)        return query(temp,x,y);    else if(x>mid)        return query(temp+1,x,y);    else        return query(temp,x,mid)+query(temp+1,mid+1,y);}void reupdate(ll t,ll x,ll y){    ll ans;   if(a[t].p)        return;    if(a[t].x==a[t].y)    {        a[t].p=1;        ans=binary(0,91,a[t].sum);        //printf("%I64d %I64d\n",a[t].sum,ans);        a[t].sum=ans;        return ;    }    ll mid=(a[t].x+a[t].y)>>1,temp=t<<1;    if(y<=mid)         reupdate(temp,x,y);    else if(x>mid)          reupdate(temp+1,x,y);    else    {        reupdate(temp,x,mid);        reupdate(temp+1,mid+1,y);    }    a[t].p=(a[temp].p)&(a[temp+1].p);    a[t].sum=a[temp].sum+a[temp+1].sum;}int main(){    fun();    ll n,m,i,op,k,d,l,r;    while(scanf("%I64d%I64d",&n,&m)!=EOF)    {        build(1,1,n);        for(i=1;i<=m;i++)        {            scanf("%I64d",&op);            if(op==1)            {                scanf("%I64d%I64d",&k,&d);                update(1,k,d);            }            else if(op==2)            {                scanf("%I64d%I64d",&l,&r);                printf("%I64d\n",query(1,l,r));            }            else            {                scanf("%I64d%I64d",&l,&r);                reupdate(1,l,r);            }        }    }    return 0;}