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.
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.
1 1 2 1 1 5 4 1 1 7 1 3 17 3 2 4 2 1 5
0 22
#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;//判断子结点是否需要进行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);//同时更新备用 return; } if(ss[d].t)//如果当前段需要进行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;//子结点需要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; }