Wow! Such sequence! The third round of the HDU multi-School Joint Competition: 1007

Wow! Such sequence!

Time Limit: 10000/5000 MS (Java/others) memory limit: 65536/65536 K (Java/Others)

Problem descriptionRecently, 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, Maid number FN is defined as fn = FN-1 + FN-2 for n ≥ 2.

Nearest Maid number of number x means the smallest FN where | fn-x | is also smallest.

Doge doesn' t believe the machine cocould respond each request in less than 10 ms. Help Doge figure out the reason.

 

InputInput 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 maid"

1 ≤ n ≤ 100000, 1 ≤ m ≤ 100000, | d | <231, all queries will be valid.

 

OutputFor each type 2 ("query sum") operation, output one line containing an integer represent the answer of this query.

 

Sample Input1 1 2 1 5 4 1 1 7 1 3 17 3 2 4 2 1 5

 

Sample output022 line segment tree bare question, maintenance whether changed to fib this mark codes:

  1 #include<set>  2 #include<cstdio>  3 #include<cstdlib>  4 #include<cstring>  5 #include<iostream>  6 #include<algorithm>  7 using namespace std;  8 const int N = 100100;  9 #define Ch1 (i<<1) 10 #define Ch2 (Ch1|1) 11 #define For(i,n) for(int i=1;i<=n;i++) 12 #define Rep(i,l,r) for(int i=l;i<=r;i++) 13  14 struct tnode{ 15     int l,r,mid; 16     long long mark,sum,sumf; 17 }T[N<<2]; 18 long long F[110]; 19 int n,m,op; 20 long long l,r; 21  22 long long ABS(long long x){ 23     if(x<0) return -x; 24     return x; 25 } 26  27 long long MAX(long long A,long long B){ 28     if(A>B) return A; 29     else    return B; 30 } 31  32 long long find(long long x){ 33     long long Min = 0 , Max = 0; 34     Rep(i,0,103) 35         if(F[i]<=x) Min = MAX(F[i],Min); 36         else if(F[i]>x){ 37             Max = F[i]; 38             break; 39         } 40     if(ABS(x-Min)<=ABS(x-Max)) return Min; 41     else                       return Max; 42 } 43  44 void Pushdown(int i){ 45     if(T[i].mark){ 46         T[Ch1].sum = T[Ch1].sumf;T[Ch1].mark = 1; 47         T[Ch2].sum = T[Ch2].sumf;T[Ch2].mark = 1; 48         T[i].mark = 0; 49     } 50 } 51  52 void Build(int l,int r,int i){ 53     T[i].l = l; T[i].r = r; T[i].mid = (l+r)>>1; 54     T[i].mark = 0;  T[i].sum = T[i].sumf = 0; 55     if(l==r){ 56         T[i].sumf = 1; 57         return; 58     } 59     Build(l,T[i].mid,Ch1);Build(T[i].mid+1,r,Ch2); 60     T[i].sumf = T[Ch1].sumf + T[Ch2].sumf;  61 } 62  63 void Modify(int i,int x,long long delta){ 64     if(T[i].l==T[i].r){ 65         T[i].sum+=delta; 66         T[i].sumf = find(T[i].sum); 67         return; 68     } 69     Pushdown(i); 70     if(x<=T[i].mid) Modify(Ch1,x,delta); 71     else            Modify(Ch2,x,delta); 72     T[i].sum = T[Ch1].sum + T[Ch2].sum; 73     T[i].sumf = T[Ch1].sumf + T[Ch2].sumf; 74 } 75  76 void Modifyf(int i,int l,int r){ 77     if(l<=T[i].l&&T[i].r<=r){ 78         T[i].mark = 1; 79         T[i].sum = T[i].sumf; 80         return; 81     } 82     Pushdown(i); 83     if(r<=T[i].mid)  Modifyf(Ch1,l,r);else 84     if(l>T[i].mid)   Modifyf(Ch2,l,r);else 85     Modifyf(Ch1,l,T[i].mid) , Modifyf(Ch2,T[i].mid+1,r); 86     T[i].sum = T[Ch1].sum + T[Ch2].sum; 87     T[i].sumf = T[Ch1].sumf + T[Ch2].sumf; 88 } 89  90 long long query(int l,int r,int i){ 91     if(l<=T[i].l&&T[i].r<=r) return T[i].sum; 92     Pushdown(i); 93     if(r<=T[i].mid) return query(l,r,Ch1);else 94     if(l>T[i].mid)  return query(l,r,Ch2);else 95     return   query(l,T[i].mid,Ch1) + query(T[i].mid+1,r,Ch2); 96      97 } 98  99 void init(){100     while(scanf("%d%d",&n,&m)!=EOF){101         Build(1,n,1);102         For(i,m){103             scanf("%d%I64d%I64d",&op,&l,&r);104             if(op==1)   Modify(1,l,r);105             if(op==2)   printf("%I64d\n",query(l,r,1));106             if(op==3)   Modifyf(1,l,r);107         }108     }109 }110 111 int main(){112     F[0] = 1; F[1] = 1;113     Rep(i,2,103) F[i] = F[i-1] + F[i-2];114     init();    115     return 0;116 }