Wow! Such Sequence!
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2234 Accepted Submission(s): 657
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 F 0 = 1,F 1 = 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.
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 F 0 = 1,F 1 = 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| < 2 31, all queries will be valid.
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| < 2 31, 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
Author
Fudan University
Source
【题目分析】
这题相比于裸的线段树区间更新有了一些难度。
我们在每个结点中设一个fib，表示离sum最近的fibnacci数，每次区间更新时，就将sum的值更新为fib。fib的值只有在单点更新的过程中才会改变，也就是说当sum值改变的时候fib才改变，因为当sum变为fib后，离sum最近的fibnacci数还是fib值。
lazy----记录该点以下的孩子结点是否需要更新。
我用long long ,然后用了%lld输入，不知道杭电不支持%lld，debug了半天，TLE了20多次，后来改%lld为%I64d就过了，哭死QAQ...
//Memory Time
// K MS
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define MAX 100010
#define LL long long
using namespace std;
LL n,m;
LL ans;
LL f[60];
struct Tree
{
LL l,r;
LL sum,fib;
bool lazy;
};
Tree tree[MAX*3];
LL Find(LL x)
{
if(x<=0)return 1;
LL ldis,rdis;
for(int i=0;i<59;++i)
{
if(f[i]<=x&&f[i+1]>=x)
{
ldis=x-f[i];
rdis=f[i+1]-x;
return ldis<=rdis?f[i]:f[i+1];
}
}
}
void pushup(LL x)
{
LL tmp=x<<1;
tree[x].sum=tree[tmp].sum+tree[tmp+1].sum;
tree[x].fib=tree[tmp].fib+tree[tmp+1].fib;
}
void pushdown(LL x)
{
if(!tree[x].lazy)return;
tree[x].lazy=0;
if(tree[x].l==tree[x].r)return;
LL tmp=x<<1;
tree[tmp].lazy=tree[tmp+1].lazy=1;
tree[tmp].sum=tree[tmp].fib;
tree[tmp+1].sum=tree[tmp+1].fib;
}
void build(LL l,LL r,LL x)
{
tree[x].l=l,tree[x].r=r;
tree[x].sum=0,tree[x].fib=1,tree[x].lazy=0;
if(l==r)return;
LL tmp=x<<1;
LL mid=(l+r)>>1;
build(l,mid,tmp);
build(mid+1,r,tmp+1);
pushup(x);
}
void add(LL x,LL k,LL num)
{
if(tree[x].l==tree[x].r)
{
tree[x].sum+=num;
tree[x].fib=Find(tree[x].sum);
return;
}
if(tree[x].lazy)
pushdown(x);
LL tmp=x<<1;
LL mid=(tree[x].l+tree[x].r)>>1;
if(k<=mid)
add(tmp,k,num);
else if(k>mid)
add(tmp+1,k,num);
pushup(x);
}
void change(LL l,LL r,LL x)
{
if(r<tree[x].l||l>tree[x].r)return;
if(l<=tree[x].l&&r>=tree[x].r)
{
tree[x].sum=tree[x].fib;
tree[x].lazy=1;
return;
}
if(tree[x].lazy)pushdown(x);
LL tmp=x<<1;
LL mid=(tree[x].l+tree[x].r)>>1;
if(r<=mid)
change(l,r,tmp);
else if(l>mid)
change(l,r,tmp+1);
else
{
change(l,mid,tmp);
change(mid+1,r,tmp+1);
}
pushup(x);
}
void query(LL l,LL r,LL x)
{
if(r<tree[x].l||l>tree[x].r)return;
if(l<=tree[x].l&&r>=tree[x].r)
{
ans+=tree[x].sum;
return;
}
if(tree[x].lazy)
pushdown(x);
LL tmp=x<<1;
LL mid=(tree[x].l+tree[x].r)>>1;
if(r<=mid)
query(l,r,tmp);
else if(l>mid)
query(l,r,tmp+1);
else
{
query(l,mid,tmp);
query(mid+1,r,tmp+1);
}
pushup(x);
}
int main()
{
// freopen("cin.txt","r",stdin);
// freopen("cout.txt","w",stdout);
f[0]=f[1]=1;
for(int i=2;i<60;++i) f[i]=f[i-1]+f[i-2];
while(scanf("%I64d %I64d",&n,&m)!=EOF)
{
build(1,n,1);
LL a,b,c;
while(m--)
{
scanf("%I64d %I64d %I64d",&a,&b,&c);
if(a==1)
add(1,b,c);
else if(a==2)
{
ans=0;
query(b,c,1);
printf("%I64d\n",ans);
}
else
change(b,c,1);
}
}
return 0;
}
