Wow! Such Sequence! HDU多校联合赛第三场1007

Wow! Such Sequence!

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

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 1

2 1 1

5 4

1 1 7

1 3 17

3 2 4

2 1 5

 

Sample Output

0

22

 

线段树裸题，维护是否改变为fib这个标记

Codes:

#include<set>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 100100;
#define Ch1 (i<<1)
#define Ch2 (Ch1|1)
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)

struct tnode{
    int l,r,mid;
    long long mark,sum,sumf;
}T[N<<2];
long long F[110];
int n,m,op;
long long l,r;

long long ABS(long long x){
    if(x<0) return -x;
    return x;
}

long long MAX(long long A,long long B){
    if(A>B) return A;
    else    return B;
}

long long find(long long x){
    long long Min = 0 , Max = 0;
    Rep(i,0,103)
        if(F[i]<=x) Min = MAX(F[i],Min);
        else if(F[i]>x){
            Max = F[i];
            break;
        }
    if(ABS(x-Min)<=ABS(x-Max)) return Min;
    else                       return Max;
}

void Pushdown(int i){
    if(T[i].mark){
        T[Ch1].sum = T[Ch1].sumf;T[Ch1].mark = 1;
        T[Ch2].sum = T[Ch2].sumf;T[Ch2].mark = 1;
        T[i].mark = 0;
    }
}

void Build(int l,int r,int i){
    T[i].l = l; T[i].r = r; T[i].mid = (l+r)>>1;
    T[i].mark = 0;  T[i].sum = T[i].sumf = 0;
    if(l==r){
        T[i].sumf = 1;
        return;
    }
    Build(l,T[i].mid,Ch1);Build(T[i].mid+1,r,Ch2);
    T[i].sumf = T[Ch1].sumf + T[Ch2].sumf; 
}

void Modify(int i,int x,long long delta){
    if(T[i].l==T[i].r){
        T[i].sum+=delta;
        T[i].sumf = find(T[i].sum);
        return;
    }
    Pushdown(i);
    if(x<=T[i].mid) Modify(Ch1,x,delta);
    else            Modify(Ch2,x,delta);
    T[i].sum = T[Ch1].sum + T[Ch2].sum;
    T[i].sumf = T[Ch1].sumf + T[Ch2].sumf;
}

void Modifyf(int i,int l,int r){
    if(l<=T[i].l&&T[i].r<=r){
        T[i].mark = 1;
        T[i].sum = T[i].sumf;
        return;
    }
    Pushdown(i);
    if(r<=T[i].mid)  Modifyf(Ch1,l,r);else
    if(l>T[i].mid)   Modifyf(Ch2,l,r);else
    Modifyf(Ch1,l,T[i].mid) , Modifyf(Ch2,T[i].mid+1,r);
    T[i].sum = T[Ch1].sum + T[Ch2].sum;
    T[i].sumf = T[Ch1].sumf + T[Ch2].sumf;
}

long long query(int l,int r,int i){
    if(l<=T[i].l&&T[i].r<=r) return T[i].sum;
    Pushdown(i);
    if(r<=T[i].mid) return query(l,r,Ch1);else
    if(l>T[i].mid)  return query(l,r,Ch2);else
    return   query(l,T[i].mid,Ch1) + query(T[i].mid+1,r,Ch2);
    
}

void init(){
    while(scanf("%d%d",&n,&m)!=EOF){
        Build(1,n,1);
        For(i,m){
            scanf("%d%I64d%I64d",&op,&l,&r);
            if(op==1)   Modify(1,l,r);
            if(op==2)   printf("%I64d
",query(l,r,1));
            if(op==3)   Modifyf(1,l,r);
        }
    }
}

int main(){
    F[0] = 1; F[1] = 1;
    Rep(i,2,103) F[i] = F[i-1] + F[i-2];
    init();    
    return 0;
}