hdu4893Wow! Such Sequence! (线段树)

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each Type 2 ("query sum") operation, output one line containing an integer represent the answer of this query.

[align=left]Sample Input[/align]

1 1
2 1 1
5 4
1 1 7
1 3 17
3 2 4
2 1 5

[align=left]Sample Output[/align]

0
22

[align=left]Source[/align]
field=problem&key=2014%20Multi-University%20Training%20Contest%203&source=1&searchmode=source">2014 Multi-University Training Contest 3

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解题：线段树结点含两个和，一个真正和，还有一个是3号操作的备用和。

#include<iostream>
#include<stdio.h>
using namespace std;
struct tree
{
int l,r,t;
__int64 s,ts;
}ss[100010*3];
__int64 f[100]={1,1};
void setit(int l,int r,int d)
{
ss[d].l=l;
ss[d].r=r;
ss[d].t=0;//推断子结点是否须要进行3号操作
ss[d].s=0;//当前区间内的总和
ss[d].ts=1;//当前区间进行3号操作的备用和
if(l==r)
return;
setit(l,(l+r)/2,d*2);
setit((l+r)/2+1,r,d*2+1);
ss[d].ts=ss[d*2].ts+ss[d*2+1].ts;
}
__int64 doublekill(__int64 k)//查找与k最相近的Fibonacci
{
int l=0,r=77,mid;
__int64 x1,x2;
while(l<=r)
{
mid=(l+r)/2;
if(f[mid]==k)return f[mid];
if(f[mid]<k)l=mid+1;
else r=mid-1;
}
x1=f[l]-k;
if(x1<0)x1=-x1;
if(l>0)
{
x2=f[l-1]-k;
if(x2<0)x2=-x2;
if(x2<=x1)return f[l-1];
}
return f[l];
}
void insert1(__int64 e,int k,int d)//点更新
{
int l,r,mid;
l=ss[d].l;
r=ss[d].r;
mid=(l+r)/2;
if(l==r)
{
if(ss[d].t)
{
ss[d].s=ss[d].ts; ss[d].t=0;
}
ss[d].s+=e; ss[d].ts=doublekill(ss[d].s);//同一时候更新备用
return;
}
if(ss[d].t)//假设当前段须要进行3号操作
{
ss[d*2].t=1; ss[d*2+1].t=1; ss[d].t=0;
ss[d*2].s=ss[d*2].ts;
ss[d*2+1].s=ss[d*2+1].ts;
}
if(k>mid)insert1(e,k,d*2+1);
else insert1(e,k,d*2);
ss[d].s=ss[d*2].s+ss[d*2+1].s;
ss[d].ts=ss[d*2].ts+ss[d*2+1].ts;
}
void insert2(int l,int r,int d)//区间进行3号操作
{
int ll,rr,mid;
ll=ss[d].l;
rr=ss[d].r;
mid=(ll+rr)/2;
if(ll>=l&&rr<=r)
{
ss[d].s=ss[d].ts; ss[d].t=1;//子结点须要3号操作更新
return;
}
if(ss[d].t)
{
ss[d*2].t=1; ss[d*2+1].t=1; ss[d].t=0;
ss[d*2].s=ss[d*2].ts;
ss[d*2+1].s=ss[d*2+1].ts;
}
if(l<=mid)insert2(l,r,d*2);
if(r>mid)insert2(l,r,d*2+1);
ss[d].s=ss[d*2].s+ss[d*2+1].s;
}
__int64 find(int l,int r,int d)//求和
{
__int64 ll,rr,mid,s=0;
ll=ss[d].l;
rr=ss[d].r;
mid=(ll+rr)/2;
if(ll>=l&&rr<=r)
{
return ss[d].s;
}
if(ss[d].t)
{
ss[d*2].t=1; ss[d*2+1].t=1; ss[d].t=0;
ss[d*2].s=ss[d*2].ts;
ss[d*2+1].s=ss[d*2+1].ts;
}
if(l<=mid)s+=find(l,r,d*2);
if(r>mid)s+=find(l,r,d*2+1);
ss[d].s=ss[d*2].s+ss[d*2+1].s;
return s;
}
int main (void)
{
int n,m,i,j,k,l;
__int64 d;
for(i=2;i<78;i++)
f[i]=f[i-1]+f[i-2];
while(scanf("%d%d",&n,&m)>0)
{
setit(1,n,1);
while(m--)
{
scanf("%d%d",&j,&k);
if(j==1)
{
scanf("%I64d",&d);insert1(d,k,1);
}
else
{
scanf("%d",&l);
if(j==2)printf("%I64d\n",find(k,l,1));
else insert2(k,l,1);
}
}
}
return 0;
}