BZOJ 3211 题解

3211: 花神游历各国

Time Limit: 5 Sec  Memory Limit: 128 MB
Submit: 2549  Solved: 946
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Description

 

Input

 

Output

每次x=1时，每行一个整数，表示这次旅行的开心度

Sample Input

4
1 100 5 5
5
1 1 2
2 1 2
1 1 2
2 2 3
1 1 4

Sample Output

101
11
11

HINT

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。

 /**************************************************************
     Problem: 3211
     User: shadowland
     Language: C++
     Result: Accepted
     Time:1568 ms
     Memory:40372 kb
 ****************************************************************/

 #include "bits/stdc++.h"

 using namespace std ;
 typedef long long QAQ ;
 struct SegTree { int l , r , sqr ; QAQ sum ; } ;
 const int maxN =  ;

 SegTree tr[ maxN <<  ] ;

 void Push_up ( const int i ) {
         tr[ i ].sum = tr[ i <<  ].sum + tr[ i <<  |  ].sum ;
 }

 int INPUT ( ) {
         int x =  , f =  ; char ch = getchar ( ) ;
         while ( ch < '' || ch > '' ) { if ( ch == '-' ) f = -  ; ch = getchar ( ) ; }
         while ( ch >= '' && ch <='' ) { x = ( x <<  ) + ( x <<  ) + ch - '' ; ch = getchar ( ) ; }
         return x * f ;
 }

 void Build_Tree ( const int x , const int y , const int i ) {
         tr[ i ].l = x ; tr[ i ].r = y ;
         if ( x == y ) tr[ i ].sum = INPUT ( ) ;
         else {
                 int mid = ( tr[ i ].l + tr[ i ].r ) >>  ;
                 Build_Tree ( x , mid , i <<  ) ;
                 Build_Tree ( mid +  , y , i <<  |  ) ;
                 Push_up ( i ) ;
         }
 }
 void Update_Tree ( const int q , const int w , const int i ) {
         if ( q <= tr[ i ].l && tr[ i ].r <= w  ) {
                 ++ tr[ i ].sqr ;
                 if ( tr[ i ].l == tr[ i ].r ) {
                         tr[ i ].sum = sqrt ( tr[ i ].sum ) ;
                         return ;
                 }
         }

         int mid = ( tr[ i ].l + tr[ i ].r ) >>  ;
         if ( tr[ i ].sqr <=  ) {
                 if ( q > mid ) Update_Tree ( q , w , i <<  |  ) ;
                 else if ( w <= mid ) Update_Tree ( q , w , i <<  ) ;
                 else {
                         Update_Tree ( q , w , i <<  |  ) ;
                         Update_Tree ( q , w , i <<  ) ;
                 }
                 Push_up ( i ) ;
         }

 }

 QAQ Query_Tree ( const int q , const int w , const int i ) {
         if ( q <= tr[ i ].l && tr[ i ].r <= w ) {
                 return tr[ i ].sum ;
         }
         else {
                 int mid = ( tr[ i ].l + tr[ i ].r ) >>  ;
                 if ( q > mid ) return Query_Tree ( q , w , i <<  | ) ;
                 else if ( w <= mid ) return Query_Tree ( q , w , i <<  ) ;
                 else return Query_Tree ( q , w , i <<  |  ) + Query_Tree ( q , w , i <<  ) ;
         }
 }

 int main ( ) {
         int N = INPUT ( ) ;
         Build_Tree (  , N ,  ) ;
         int Q = INPUT( ) ;
         while ( Q-- ) {
                 int op = INPUT ( ) ;
                 if( op ==  ) {
                         int l = INPUT( ) , r = INPUT( ) ;
                         printf ( "%lld\n" , Query_Tree ( l , r ,  ) ) ;
                 }
                 else if ( op ==  ) {
                         int l = INPUT ( ) , r = INPUT ( ) ;
                         Update_Tree ( l , r ,  ) ;
                 }
         }
         return  ;
 }

2016-10-13 22:10:52

(完)