【ZOJ】3886 Nico Number【线段树】

传送门：【ZOJ】3886 Nico Number

YY方法出只有2的幂次和素数是等差的，打表验证发现只有6是反例，注意到0项也是等差，所以0和1也算。每个数最多logn次取模，所以区间更新暴力到叶子就行。ZOJ月赛的时候抢了这题的FB ^_^

my  code:

#include <bits/stdc++.h>
using namespace std ;

typedef long long LL ;
typedef long long Int ;
typedef pair < int , int > pi ;

#define clr(a,x) memset ( a , x , sizeof a ) 
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define root 1 , 1 , n
#define mid ( ( l + r ) >> 1 )

const int MAXN = 100005 ;
const int MAXM = 10000005 ;

bool prime[MAXM] ;
int pri[1000005] , cnt ;
int sum[MAXN << 2] ;
int maxv[MAXN << 2] ;
int n , m ;
int S[MAXN] , top ;

int gcd ( int a , int b ) {
    return b ? gcd ( b , a % b ) : a ;
}

void preprocess () {
    for ( int i = 2 ; i < MAXM ; ++ i ) {
        if ( !prime[i] ) pri[cnt ++] = i ;
        for ( int j = 0 ; j < cnt && i * pri[j] < MAXM ; ++ j ) {
            prime[i * pri[j]] = 1 ;
            if ( i % pri[j] == 0 ) break ;
        }
    }
    for ( int i = 2 ; i < MAXM ; i <<= 1 ) prime[i] = 0 ;
    prime[6] = 0 ;
}

void pushup ( int o ) {
    sum[o] = sum[ls] + sum[rs] ;
    maxv[o] = max ( maxv[ls] , maxv[rs] ) ;
}

void build ( int o , int l , int r ) {
    if ( l == r ) {
        int x ;
        scanf ( "%d" , &x ) ;
        sum[o] = !prime[x] ;
        maxv[o] = x ;
        return ;
    }
    int m = mid ;
    build ( lson ) ;
    build ( rson ) ;
    pushup ( o ) ;
}

void update ( int L , int R , int v , int o , int l , int r ) {
    if ( maxv[o] < v ) return ;
    if ( l == r ) {
        maxv[o] = maxv[o] % v ;
        sum[o] = !prime[maxv[o]] ;
        return ;
    }
    int m = mid ;
    if ( L <= m ) update ( L , R , v , lson ) ;
    if ( m <  R ) update ( L , R , v , rson ) ;
    pushup ( o ) ;
}

void modify ( int x , int v , int o , int l , int r ) {
    if ( l == r ) {
        maxv[o] = v ;
        sum[o] = !prime[v] ;
        return ;
    }
    int m = mid ;
    if ( x <= m ) modify ( x , v , lson ) ;
    else modify ( x , v , rson ) ;
    pushup ( o ) ;
}

int query ( int L , int R , int o , int l , int r ) {
    if ( L <= l && r <= R ) return sum[o] ;
    int m = mid ;
    if ( R <= m ) return query ( L , R , lson ) ;
    if ( m <  L ) return query ( L , R , rson ) ;
    return query ( L , R , lson ) + query ( L , R , rson ) ;
}

void solve () {
    int op , L , R , v ;
    build ( root ) ;
    scanf ( "%d" , &m ) ;
    for ( int i = 1 ; i <= m ; ++ i ) {
        scanf ( "%d" , &op ) ;
        if ( op == 1 ) {
            scanf ( "%d%d" , &L , &R ) ;
            int ans = query ( L , R , root ) ;
            printf ( "%d\n" , ans ) ;
        } else if ( op == 2 ) {
            scanf ( "%d%d%d" , &L , &R , &v ) ;
            update ( L , R , v , root ) ;
        } else {
            scanf ( "%d%d" , &L , &v ) ;
            modify ( L , v , root ) ;
        }
    }
}

int main () {
    preprocess () ;
    while ( ~scanf ( "%d" , &n ) ) solve () ;
    return 0 ;
}