BZOJ 3211 题解

对于100%的数据， n ≤ 100000，m≤200000 ,data[i]非负且小于10^9

Solution

区间开根号看似十分难处理，实际不妨找规律。data[ i ] 的值最大为10⁹。

( 10⁹)^½= 10⁵

( ( 10⁹)^½)^½ = 10³

( ( ( 10⁹)^½)^½ )^½ = 10²

( ( ( ( 10⁹)^½)^½ )^½ )^½ = 10¹

( ( ( ( ( 10⁹)^½)^½ )^½ )^½)^½ = 1

一个很大的数，五次根号后全部是1，当某个区间的值被开过5次根号，那么该区间所有值都是1.

根据以上结论，不难写出代码。

 1 /**************************************************************
 2     Problem: 3211
 3     User: shadowland
 4     Language: C++
 5     Result: Accepted
 6     Time:1568 ms
 7     Memory:40372 kb
 8 ****************************************************************/
 9 
10 #include "bits/stdc++.h"
11  
12 using namespace std ;
13 typedef long long QAQ ;
14 struct SegTree { int l , r , sqr ; QAQ sum ; } ; 
15 const int maxN = 500100 ;
16 
17 SegTree tr[ maxN << 2 ] ;
18 
19 void Push_up ( const int i ) {
20         tr[ i ].sum = tr[ i << 1 ].sum + tr[ i << 1 | 1 ].sum ;
21 }
22 
23 int INPUT ( ) {
24         int x = 0 , f = 1 ; char ch = getchar ( ) ;
25         while ( ch < '0' || ch > '9' ) { if ( ch == '-' ) f = - 1 ; ch = getchar ( ) ; }
26         while ( ch >= '0' && ch <='9' ) { x = ( x << 1 ) + ( x << 3 ) + ch - '0' ; ch = getchar ( ) ; }
27         return x * f ;
28 }
29 
30 void Build_Tree ( const int x , const int y , const int i ) {
31         tr[ i ].l = x ; tr[ i ].r = y ;
32         if ( x == y ) tr[ i ].sum = INPUT ( ) ;
33         else {
34                 int mid = ( tr[ i ].l + tr[ i ].r ) >> 1 ;
35                 Build_Tree ( x , mid , i << 1 ) ;
36                 Build_Tree ( mid + 1 , y , i << 1 | 1 ) ;
37                 Push_up ( i ) ;
38         }
39 }
40 void Update_Tree ( const int q , const int w , const int i ) {
41         if ( q <= tr[ i ].l && tr[ i ].r <= w  ) {
42                 ++ tr[ i ].sqr ;
43                 if ( tr[ i ].l == tr[ i ].r ) {
44                         tr[ i ].sum = sqrt ( tr[ i ].sum ) ;
45                         return ;
46                 }
47         }
48         
49         int mid = ( tr[ i ].l + tr[ i ].r ) >> 1 ;
50         if ( tr[ i ].sqr <= 5 ) {
51                 if ( q > mid ) Update_Tree ( q , w , i << 1 | 1 ) ;
52                 else if ( w <= mid ) Update_Tree ( q , w , i << 1 ) ;
53                 else {
54                         Update_Tree ( q , w , i << 1 | 1 ) ;
55                         Update_Tree ( q , w , i << 1 ) ;
56                 }
57                 Push_up ( i ) ;
58         }
59         
60 }
61 
62 QAQ Query_Tree ( const int q , const int w , const int i ) {
63         if ( q <= tr[ i ].l && tr[ i ].r <= w ) {
64                 return tr[ i ].sum ;
65         }
66         else {
67                 int mid = ( tr[ i ].l + tr[ i ].r ) >> 1 ;
68                 if ( q > mid ) return Query_Tree ( q , w , i << 1 | 1) ;
69                 else if ( w <= mid ) return Query_Tree ( q , w , i << 1 ) ;
70                 else return Query_Tree ( q , w , i << 1 | 1 ) + Query_Tree ( q , w , i << 1 ) ;
71         }
72 }
73 
74 int main ( ) {
75         int N = INPUT ( ) ;
76         Build_Tree ( 1 , N , 1 ) ;
77         int Q = INPUT( ) ;
78         while ( Q-- ) {
79                 int op = INPUT ( ) ;
80                 if( op == 1 ) {
81                         int l = INPUT( ) , r = INPUT( ) ;
82                         printf ( "%lld
" , Query_Tree ( l , r , 1 ) ) ;
83                 } 
84                 else if ( op == 2 ) {
85                         int l = INPUT ( ) , r = INPUT ( ) ;
86                         Update_Tree ( l , r , 1 ) ;
87                 }
88         }
89         return 0 ;
90 }

View Code

2016-10-13 22:10:52

(完)

3211: 花神游历各国

Description

Input

Output

Sample Input

Sample Output

HINT

Solution