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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 Input1 1 2 1 1 5 4 1 1 7 1 3 17 3 2 4 2 1 5
Sample Output022 線段樹裸題,維護是否改變為fib這個標記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 }