hdu 4893 Wow! Such Sequence!（线段树和树状数组实现）--2014 Multi-University Training Contest 3

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4893

Wow! Such Sequence!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 435    Accepted Submission(s): 133

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

 

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一道线段树的问题，当然也可以用树状数组实现，难点主要在把一个区间的数改成离自己最近的斐波那契数的那一块。需要进行优化，我的做法是通过set记录，已经处理过的节点，避免一个节点被重复处理，因为重复处理的结果肯定是一样的。

线段树实现：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <string>#include <bitset>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <cstdlib>using namespace std;#define CLR(A) memset(A,0,sizeof(A))typedef long long ll;const int MAX=400010;struct Node{    int l,r;    ll v;}tree[MAX];ll f[100];int del[MAX];void build(int id,int l,int r){    tree[id].l=l;tree[id].r=r;tree[id].v=0;    if(l==r) return;    int m=(l+r)>>1;    build(id<<1,l,m);    build(id<<1|1,m+1,r);}void add(int id,int i,ll v){    if(tree[id].l==tree[id].r){        tree[id].v+=v;        return;    }    int m=(tree[id].l+tree[id].r)>>1;    if(i<=m) add(id<<1,i,v);    else add(id<<1|1,i,v);    tree[id].v=tree[id<<1].v+tree[id<<1|1].v;}ll query(int id,int l,int r){    if(l==tree[id].l&&r==tree[id].r) return tree[id].v;    int m=(tree[id].l+tree[id].r)>>1;    if(r<=m) return query(id<<1,l,r);    if(l>m) return query(id<<1|1,l,r);    return query(id<<1,l,m)+query(id<<1|1,m+1,r);}void init(){    CLR(f);    f[0]=1;f[1]=1;    for(int i=2;i<=91;i++) f[i]=f[i-1]+f[i-2];}int main(){    int n,m;    init();    while(~scanf("%d%d",&n,&m)){        build(1,1,n);        set<int>st;        set<int>::iterator it;        for(int i=1;i<=n;i++) st.insert(i);        while(m--){            int ord,a,b;            scanf("%d%d%d",&ord,&a,&b);            switch(ord){                case 1:                    add(1,a,b);                    st.insert(a);                    break;                case 2:                    printf("%I64d\n",query(1,a,b));                    break;                case 3:                    it=lower_bound(st.begin(),st.end(),a);                    del[0]=0;                    for(;it!=st.end()&&((*it)<=b);it++){                        int p=*it;                        del[++del[0]]=p;                        ll u=query(1,p,p),k;                        int t=lower_bound(f+1,f+92,u)-f;                        if(abs(f[t]-u)<abs(f[t-1]-u)) k=f[t];                        else k=f[t-1];                        add(1,p,-u);add(1,p,k);                    }                    for(int i=1;i<=del[0];i++) st.erase(del[i]);                    break;                default:;            }        }    }    return 0;}

树状态数组实现：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <string>#include <bitset>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <cstdlib>using namespace std;#define CLR(A) memset(A,0,sizeof(A))typedef long long ll;const int MAX=100010;ll bit[MAX];ll f[100];int del[MAX];int n,m;inline int lowbit(int t){return t&-t;}ll sum(int i){    ll s=0;    while(i>0){        s+=bit[i];        i-=lowbit(i);    }    return s;}void add(int i,ll v){    while(i<=n){        bit[i]+=v;        i+=lowbit(i);    }}void init(){    CLR(f);    f[0]=1;f[1]=1;    for(int i=2;i<=91;i++) f[i]=f[i-1]+f[i-2];}int main(){    init();    while(~scanf("%d%d",&n,&m)){        CLR(bit);        set<int>st;        set<int>::iterator it;        for(int i=1;i<=n;i++) st.insert(i);        while(m--){            int ord,a,b;            scanf("%d%d%d",&ord,&a,&b);            switch(ord){                case 1:                    add(a,b);                    st.insert(a);                    break;                case 2:                    printf("%I64d\n",sum(b)-sum(a-1));                    break;                case 3:                    it=lower_bound(st.begin(),st.end(),a);                    del[0]=0;                    for(;it!=st.end()&&((*it)<=b);it++){                        int p=*it;                        del[++del[0]]=p;                        ll u=sum(p)-sum(p-1),k;                        int t=lower_bound(f+1,f+92,u)-f;                        if(abs(f[t]-u)<abs(f[t-1]-u)) k=f[t];                        else k=f[t-1];                        add(p,-u);add(p,k);                    }                    for(int i=1;i<=del[0];i++) st.erase(del[i]);                    break;                default:;            }        }    }    return 0;}