Wow! Such Sequence! Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2512    Accepted Submission(s): 751


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




 


Author


Fudan University


 


Source






题意不说了，非常easy的线段树题目了。就是打标记改点求段，改动段时因为极限次数不多。直接暴力更新到点，


好久没写线段树了，错了好多次，写的好傻逼。




代码：




/* ***********************************************
Author :rabbit
Created Time :2014/8/4 14:58:15
File Name :11.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
ll fib[100];
struct node{
    ll l,r;
    ll sum,flag;
}a[800300];
ll m,n;
void pushup(ll t){
    if(a[t].l==a[t].r)return;
    a[t].sum=a[2*t].sum+a[2*t+1].sum;
    a[t].flag=a[2*t].flag&a[2*t+1].flag;
}
void build(ll t,ll l,ll r){
//    cout<<"hhh "<<l<<" "<<r<<endl;
    a[t].l=l;
    a[t].r=r;
    a[t].sum=a[t].flag=0;
    if(l==r)return;
    ll mid=(l+r)/2;
    build(2*t,l,mid);
    build(2*t+1,mid+1,r);
    pushup(t);
}
void update1(ll t,ll p,ll val){
    if(a[t].l==a[t].r){
        a[t].sum+=val;
        a[t].flag=0;
        return;
    }
    ll mid=(a[t].l+a[t].r)/2;
    if(p<=mid)update1(2*t,p,val);
    else update1(2*t+1,p,val);
    pushup(t);
}
ll Find(ll x){
    if(x <= 1)return 1;
    int l = 1, r = 80, id = 80;
    while(l <= r){
    int mid = l+r>>1;
        if(fib[mid] > x)    id = mid, r = mid-1;
        else    l = mid+1;
    }
    if(x-fib[id-1] <= fib[id]-x)    return fib[id-1];
    return fib[id];
}

void update2(ll t,ll l,ll r){
    if(a[t].flag)return;
    if(a[t].l==a[t].r){
        a[t].sum=Find(a[t].sum);
        a[t].flag=1;
        //cout<<"ddd "<<l<<" "<<r<<" "<<a[t].sum<<endl;
        return;
    }
    ll mid=(a[t].l+a[t].r)/2;
    if(l<=mid)update2(2*t,l,r);
    if(r>mid)update2(2*t+1,l,r);
    pushup(t);
}

ll getsum(ll t,ll l,ll r){
    if(a[t].l>=l&&a[t].r<=r)return a[t].sum;
    ll mid=(a[t].l+a[t].r)/2;
    ll ans=0;
    if(l<=mid)ans+=getsum(2*t,l,r);
    if(r> mid) ans+=getsum(2*t+1,l,r);
    return ans;
}
int main()
{
    // freopen("data.in","r",stdin);
    // freopen("data.out","w",stdout);
     fib[0]=1;fib[1]=1;
     for(ll i=2;i<=90;i++)fib[i]=fib[i-1]+fib[i-2];
     while(~scanf("%I64d%I64d",&n,&m)){
        // cout<<"ddd "<<endl;
         build(1,1,n);
         //cout<<"ppp "<<endl;
         while(m--){
             ll l,r,op;
             scanf("%I64d%I64d%I64d",&op,&l,&r);
             if(op==1){
                 update1(1,l,r);
            //     cout<<"han 1"<<endl;
             }
              if(op==2){
                 printf("%I64d\n",getsum(1,l,r));
                // cout<<"han 2"<<endl;
             }
              if(op==3){
                 update2(1,l,r);
            //     cout<<"han 3"<<endl;
             }
         }
     }
     return 0;
}
/*
5 10 
2 1 5 
3 1 5 
2 1 5 
1 1 10 
2 1 5 
3 1 5 
2 1 5 
 
4 5 
1 1 3 
2 1 2 
3 2 3 
1 2 1 
2 1 4 
*/