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): 976    Accepted Submission(s): 298

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


对于每个点距离它最近的Fibonacci数需要加多少，那么就知道一个区间需要加多少了。。
const int M=90;
const int N=100005;
ll f[100],sum[N<<2],need[N<<2];
bool lazy[N<<2];

inline ll Find(ll x){
    if(x<=1) return 1;
    int id=lower_bound(f,f+M,x)-f;
    if(x-f[id-1]<=f[id]-x) return f[id-1];
    return f[id];
}

inline void push_up(int t){
    sum[t]=sum[L]+sum[R];
    need[t]=need[L]+need[R];
}

inline void push_down(int t){
    if(lazy[t]){
        lazy[L]=lazy[R]=1;
        sum[L]+=need[L];
        sum[R]+=need[R];
        need[L]=need[R]=0;
        lazy[t]=0;
    }
}

inline void build(int t,int l,int r){
    sum[t]=0;lazy[t]=0;
    need[t]=(r-l+1);
    if(l!=r){
        MID(l,r);
        build(L,l,mid);
        build(R,mid+1,r);
    }
}

inline void add(int t,int l,int r,int x,ll v){
    if(l==r){
        sum[t]+=v;
        need[t]=Find(sum[t])-sum[t];
        return;
    }
    push_down(t);
    MID(l,r);
    if(x<=mid) add(L,l,mid,x,v);
    else add(R,mid+1,r,x,v);
    push_up(t);
}

inline void change(int t,int l,int r,int x,int y){
    if(l>=x && r<=y){
        lazy[t]=1;
        sum[t]+=need[t];
        need[t]=0;
        return;
    }
    push_down(t);
    MID(l,r);
    if(y<=mid) change(L,l,mid,x,y);
    else if(x>mid) change(R,mid+1,r,x,y);
    else{
        change(L,l,mid,x,mid);change(R,mid+1,r,mid+1,y);
    }
    push_up(t);
}

inline ll query(int t,int l,int r,int x,int y){
    if(l>=x && r<=y) return sum[t];
    push_down(t);
    MID(l,r);
    if(y<=mid) return query(L,l,mid,x,y);
    else if(x>mid) return query(R,mid+1,r,x,y);
    else return query(L,l,mid,x,mid)+query(R,mid+1,r,mid+1,y);
}

int n,m;
int op,x,y;
int main(){
    f[0]=f[1]=1;
    rep(i,2,M) f[i]=f[i-1]+f[i-2];
    while(~scanf("%d%d",&n,&m)){
        build(1,1,n);
        while(m--){
            op=input();
            if(op==1){
                x=input(),y=input();
                add(1,1,n,x,y*1LL);
            }
            else if(op==3){
                x=input(),y=input();
                change(1,1,n,x,y);
            }
            else{
                x=input(),y=input();
                output(query(1,1,n,x,y));
            }
        }
    }
    return 0;
}