Solution to a problem - not the normal sequence gang + State +
Background decent praise
on ssw02, he bombed
Abnormal gang
The main idea
50 minutes practice
Open barrels of violence, statistics can be.
60 minutes practice
Chairman of the tree plus violence
100 practice
50 points available from the practice squad offline Mo click on it. (By the way, here ZYC seniors code optimization a bit ssw02 his team Mo board)
Can be obtained from the 60 points, the President of trees recorded on a location.
Segment tree approach is also recorded on a location. (These methods range dyeing it a)
#include<bits/stdc++.h>
using namespace std ;
#define ll long long
const int MAXN = 100005 ;
inline int read(){
int s=0 , w=1 ; char g=getchar() ;while(g>'9'||g<'0'){if(g=='-')w=-1;g=getchar();}
while(g>='0'&&g<='9')s=s*10+g-'0',g=getchar() ;return w*s ;
}
struct Seg{
int l , r , id ;
}t[ MAXN ];
int N , M , block , a[ MAXN ] , pos[ MAXN ] , num[ MAXN ] ;
ll tot = 0 , ans[ MAXN ] ;
inline bool cmp( Seg x , Seg y ){
return ( pos[ x.l ]==pos[ y.l ] )?( x.r < y.r ) : ( pos[x.l] < pos[y.l] ) ;
}
inline void updata( int x , int opt ){
if( !opt ){//收缩 扩张
--num[ a[ x ] ] ;
if( num[ a[ x ] ] > 2 ) tot -= a[ x ];
else if(num[ a[ x ] ] == 2) tot -= a[ x ]*3 ;
else if(num[ a[ x ] ] == 1) tot += a[ x ] ;
else tot -= a[ x ] ;
return;
}
++num[ a[ x ] ] ;
if( num[ a[ x ] ] > 3 ) tot += a[ x ];
else if( num[ a[ x ] ] == 3 ) tot += a[ x ]*3 ;
else if( num[ a[ x ] ] == 2 ) tot -= a[ x ] ;
else tot += a[ x ];
}
void Mo(){
for( int l = 1 , r = 0 , i = 1 ; i <= M ; ++i ){
while( l < t[ i ].l )updata( l++ , 0 ) ;
while( l > t[ i ].l )updata( --l , 1 ) ;
while( r < t[ i ].r )updata( ++r , 1 ) ;
while( r > t[ i ].r )updata( r-- , 0 ) ;
ans[ t[ i ].id ] = tot ;
}
}
int main(){
freopen("abnormal.in","r",stdin);
freopen("abnormal.out","w",stdout);
N = read() , M = read() , block = sqrt(N) ;
for( int i = 1 ; i <= N ; ++i )a[ i ] = read() ;
block = sqrt( N ) ;
for( int i = 1 ; i <= N ; ++i )pos[ i ] = (i-1) / block+1 ;
for( int i = 1 ; i <= M ; ++i )
t[ i ].l = read() , t[ i ].r = read() , t[ i ].id = i ;
sort( t+1 , t+M+1 , cmp ) ;
Mo() ;
for( int i = 1 ; i <= M ; ++i )printf("%lld\n",ans[ i ] ) ;
return 0 ;
} As a OIer, to keep the topic of respect for people, especially school seniors gold out of the question.
Not a normal country
Trie tree split heavy chain + + heuristic merge (SXK, STD)
Trie tree delete + No heuristic merge (ZYC)
tree sectional + + heuristic merge Trie tree (in fact, the first and almost)
Not the normal sequence
Cancer of the topic and people all perfectly balanced tree card. (Mang in the past when there hzy last card)
Seeking median, with a double stack it. Smaller than a constant level.
#include<bits/stdc++.h>
using namespace std ;
#define ll long long
const int MAXN = 1000005 , mod = 1e9+7 ;
inline ll read(){
ll s=0 ; char g=getchar() ; while(g>'9'||g<'0')g=getchar() ;
while( g>='0'&&g<='9')s=s*10+g-'0',g=getchar() ; return s ;
}
ll ans = 0 ;
ll a[ MAXN ] , N , A , B , C ;
priority_queue<ll>ma,mi ;
int deal( int N ){
if( N%2 )return N/2+1 ;
else return N/2 ;
}
void insert( int x ){
int u = mi.top() ;
if( x < u )mi.push( x ) ;
else ma.push( -x ) ;
while( mi.size() < ma.size() ){
int u = -ma.top() ; ma.pop() ;
mi.push( u ) ;
}
while( mi.size()-1 > ma.size() ){
int u = mi.top() ; mi.pop() ;
ma.push( -u ) ;
}
}
int main(){
freopen("unnormal.in","r",stdin) ;
freopen("unnormal.out","w",stdout) ;
A = read() , B = read() , C = read() , N = read() ;
mi.push(1) ;
ans = 1LL ;
for( ll i = 2 ; i <= N ; ++i ){
ll op = mi.top() ;
op = ( op*A%mod + B*i%mod + C )%mod ;
insert(op);ans += (ll)op ;
}
cout<<ans<<endl ;
}