bzoj 3211 线段树

开方操作最多进行5次就可以把出现的任何数变成1.

所以用线段树暴力修改,以后修改时只需看一下是否当前区间都是0或1,如果是那么就直接返回.

 /**************************************************************
     Problem: 3211
     User: idy002
     Language: C++
     Result: Accepted
     Time:1976 ms
     Memory:7068 kb
 ****************************************************************/

 #include <cstdio>
 #include <cmath>
 #define N 100010

 typedef long long dnt;
 struct Node {
     dnt s;
     bool t;
     Node *ls, *rs;
 }pool[N*], *tail=pool, *root;

 int n, m;
 int aa[N];

 Node *build( int lf, int rg ) {
     Node *nd = ++tail;
     if( lf==rg ) {
         nd->s = aa[lf];
         nd->t = aa[lf]<=;
         return nd;
     }
     int mid=(lf+rg)>>;
     nd->ls = build( lf, mid );
     nd->rs = build( mid+, rg );
     nd->s = nd->ls->s + nd->rs->s;
     nd->t = nd->ls->t && nd->rs->t;
     return nd;
 }
 void modify( Node *nd, int lf, int rg, int L, int R ) {
     if( lf==rg ) {
         nd->s = (dnt)sqrt(nd->s);
         nd->t = nd->s<=;
         return;
     }
     if( nd->t ) return;
     int mid=(lf+rg)>>;
     if( L<=mid ) modify( nd->ls, lf, mid, L, R );
     if( R>mid ) modify( nd->rs, mid+, rg, L, R );
     nd->s = nd->ls->s + nd->rs->s;
     nd->t = nd->ls->t && nd->rs->t;
 }
 dnt query( Node *nd, int lf, int rg, int L, int R ) {
     if( L<=lf && rg<=R ) return nd->s;
     int mid=(lf+rg)>>;
     dnt rt = ;
     if( L<=mid ) rt+=query( nd->ls, lf, mid, L, R );
     if( R>mid ) rt+=query( nd->rs, mid+, rg, L, R );
     return rt;
 }

 int main() {
     scanf( "%d", &n );
     for( int i=; i<=n; i++ )
         scanf( "%d", aa+i );
     root = build( , n );
     scanf( "%d", &m );
     for( int i=,opt,l,r; i<=m; i++ ) {
         scanf( "%d%d%d", &opt, &l, &r );
         if( opt== )
             printf( "%lld\n", query(root,,n,l,r) );
         else
             modify(root,,n,l,r);
     }
 }