A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help.

You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line.

Notice that the square root operation should be rounded down to integer.

题意：有一列数，共两个操作，将区间每个数开成平方根（取四舍五入），和求区间和。

线段树，由于四舍五入开根，最后一定会开成1，所以记录区间和以及区间内是否全是1，若全是1则不需要再进行开根操作了。

1 #include
      
        2 #include
       
         3 #include
        
          4 using namespace std; 5 typedef long long ll; 6 const int maxm=100005; 7  8 ll st[maxm<<2],a[maxm]; 9 bool ch[maxm<<2];10 int ql,qr;11 12 void build(int o,int l,int r){13     if(l==r){14         st[o]=a[l];15         if(st[o]==1)ch[o]=1;16         else ch[o]=0;17         return;18     }19     int m=l+((r-l)>>1);20     build(o<<1,l,m);21     build(o<<1|1,m+1,r);22     st[o]=st[o<<1]+st[o<<1|1];23     ch[o]=ch[o<<1]&ch[o<<1|1];24 }25 26 void update(int o,int l,int r){27     if(ch[o]==1)return;28     if(l==r){29         st[o]=sqrt(st[o]*1.0);30         if(st[o]==1)ch[o]=1;31         else ch[o]=0;32         return;33     }34     int m=l+((r-l)>>1);35     if(ql<=m)update(o<<1,l,m);36     if(qr>=m+1)update(o<<1|1,m+1,r);37     st[o]=st[o<<1]+st[o<<1|1];38     ch[o]=ch[o<<1]&ch[o<<1|1];39 }40 41 ll query(int o,int l,int r){42     if(ql<=l&&qr>=r)return st[o];43     int m=l+((r-l)>>1);44     ll ans=0;45     if(ql<=m)ans+=query(o<<1,l,m);46     if(qr>=m+1)ans+=query(o<<1|1,m+1,r);47     return ans;48 }49 50 int main(){51     int n;52     int cnt=0;53     while(scanf("%d",&n)!=EOF){54         int i;55         for(i=1;i<=n;++i)scanf("%I64d",&a[i]);56         build(1,1,n);57         int m;58         scanf("%d",&m);59         printf("Case #%d:\n",++cnt);60         for(i=1;i<=m;++i){61             int f;62             scanf("%d%d%d",&f,&ql,&qr);63             if(ql>qr){64                 int t=ql;ql=qr;qr=t;65             }66             if(f==1)printf("%I64d\n",query(1,1,n));67             else update(1,1,n);68         }69         printf("\n");70     }71     return 0;72 }