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Wow! Such Sequence! HDU多校联合赛第三场1007

Wow! Such Sequence!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)


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 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.
 

 

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

 

Sample Output
0
22
 
线段树裸题，维护是否改变为fib这个标记
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 }

 

posted @ 2014-07-29 16:50  ZJDx1998  阅读(238)  评论(0)  编辑  收藏  举报