HDU 4893 線段樹裸題

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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 DescriptionRecently, 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. 
InputInput 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. 
OutputFor 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

 
AuthorFudan University 
Source2014 Multi-University Training Contest 3

題意不說了，很簡單的線段樹題目了，就是打標記改點求段，修改段時由於極限次數不多，直接暴力更新到點，

好久沒寫線段樹了，錯了好多次，寫的好傻逼。

代碼：

/* ***********************************************Author :rabbitCreated Time :2014/8/4 14:58:15File 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 */