hdu 4027 Can you answer these queries? (区间线段树，区间数开方与求和，经典题目)

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 10249    Accepted Submission(s): 2350

Problem Description

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.

 

Input

The input contains several test cases, terminated by EOF.
  For
each test case, the first line contains a single integer N, denoting
there are N battleships of evil in a line. (1 <= N <= 100000)
  The
second line contains N integers Ei, indicating the endurance value of
each battleship from the beginning of the line to the end. You can
assume that the sum of all endurance value is less than 2⁶³.
  The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000)
  For
the following M lines, each line contains three integers T, X and Y.
The T=0 denoting the action of the secret weapon, which will decrease
the endurance value of the battleships between the X-th and Y-th
battleship, inclusive. The T=1 denoting the query of the commander which
ask for the sum of the endurance value of the battleship between X-th
and Y-th, inclusive.

 

Output

For
each test case, print the case number at the first line. Then print one
line for each query. And remember follow a blank line after each test
case.

 

Sample Input

10
1 2 3 4 5 6 7 8 9 10
5
0 1 10
1 1 10
1 1 5
0 5 8
1 4 8 

 

Sample Output

Case #1:
19
7
6

Source

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

之前听说过很多神级线段树的题，最后都是暴力拼循环节，今天难得见一个入门级别的，也算涨姿势了～

题意：给定n个数（1～100000），两种操作（1～100000）0 or 1 ,0操作是把给定区间内的数都变成原来的开方（向下取整），1操作，询问区间内的数的总和；

突破点：所有数据和的范围是2的63此方，大神们发现开方六七次以后，这些数就都变成1了～所以就不需要再向下更新了，

那么判定不需要更新的条件就是：该节点下的区间总和与该区间长度相等，那么它们妥妥都是1了，就可以返回。

但是本题坑坑比较多～

1坑：给定区间 Xth 与 Yth大小关系不定，记得用下swap~，不然就是RE（非法访问）。

2坑 : 每组测试实例最后还需要多输出一个空行～

0.0也怪不细心～

 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>
 #include <math.h>
 #include <iostream>
 #include <algorithm>
 #include <climits>
 #include <queue>
 #define ll long long

 using namespace std;

 const int MAX = ;
 ll num[MAX];

 struct nodes
 {
     int left,right;
     ll large;
 } tree[MAX*];

 void pushup(int root)
 {
     tree[root].large = tree[root*].large + tree[root*+].large;
 }
 void build(int root,int left,int right)
 {
     tree[root].left = left;
     tree[root].right = right;
     if(left == right)
     {
          tree[root].large = num[left];
          return;
     }

    int mid = (left+right)/;

     build(*root,left,mid);
     build(*root+,mid+,right);

     pushup(root);
 }

 void update(int root,int left,int right)
 {
     if(tree[root].large == tree[root].right - tree[root].left + )
         return;
     if(tree[root].right == tree[root].left)
        {
            tree[root].large = (ll)sqrt(tree[root].large);
            return;
        }
     int mid = (tree[root].left+tree[root].right)/;
     if(left <= mid && right > mid)
     {
         update(*root,left,mid);
         update(*root+,mid+,right);
     }
     else if(left > mid)
     {
         update(*root+,left,right);
     }
     else
     {
         update(*root,left,right);
     }
     pushup(root);
 }

 ll query(int root ,int left,int right)
 {
     if(tree[root].large == tree[root].right - tree[root].left + )
     {
         return right - left + ;
     }
     if(left == tree[root].left && right == tree[root].right)
     {
         return  tree[root].large;
     }
     int mid = (tree[root].left+tree[root].right)/;
     ll ans = ;
     if(right > mid && left <= mid)
     {
         ans += query(*root,left,mid);
         ans += query(*root+,mid+,right);
     }
     else if(left > mid)
     {
         ans += query(*root+,left,right);
     }
     else
     {
         ans += query(*root,left,right);
     }
     return ans;
 }

 int main(void)
 {
     int i,cmd,x,y,n,q,cnt= ;
     while(scanf("%d",&n) != -)
     {
         for(i = ; i <= n; i++)
             scanf("%lld",&num[i]);
             build(,,n);
         scanf("%d",&q);
         printf("Case #%d:\n",cnt++);
         for(i = ; i <q; i++)
         {
             scanf("%d %d %d",&cmd,&x,&y);
             if(y < x)
                 swap(x,y);
             if(cmd)
                 printf("%lld\n",query(,x,y));
             else
                 update(,x,y);
         }
         printf("\n");
     }
     return ;
 }

精简版

 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>
 #include <math.h>
 #include <iostream>
 #include <algorithm>
 #include <climits>
 #include <queue>
 #define ll long long

 using namespace std;

 struct Segment_T
 {
     static const int MAX = 1e5+;
     ll num[MAX];
     struct nodes
     {
         int lt,rt;
         ll add,sum;
     } T[MAX*];

     Segment_T(){}

     void pushup(int R)
     {
         T[R].sum = T[R<<].sum + T[R<<|].sum;
     }
     void pushdown(int R)
     {
         if(T[R].add)
         {
             T[R<<].add += T[R].add;
             T[R<<|].add += T[R].add;
             T[R<<].sum += T[R].add * (T[R<<].rt - T[R<<].lt + );
             T[R<<|].sum += T[R].add * (T[R<<|].rt - T[R<<|].lt + );
             T[R].add = ;
         }
     }
     void build(int R,int lt,int rt)
     {
         T[R].lt = lt;
         T[R].rt = rt;
         if(lt == rt)
         {
             T[R].sum = num[lt];
             return;
         }
         int mid = (lt+rt)>>;

         build(R<<,lt,mid);
         build(R<<|,mid+,rt);

         pushup(R);
     }
     void update(int R,int lt,int rt)
     {
         if(T[R].sum == T[R].rt - T[R].lt + )
             return;
         if(T[R].rt == T[R].lt)
         {
             T[R].sum = (ll)sqrt(T[R].sum); //返回操作处
             return;
         }
         int mid = (T[R].lt+T[R].rt)>>;
         if(lt <= mid && rt > mid)
         {
             update(R<<,lt,mid);
             update(R<<|,mid+,rt);
         }
         else if(lt > mid)
             update(R<<|,lt,rt);
         else
             update(R<<,lt,rt);
         pushup(R);
     }
     ll query(int R,int lt,int rt)
     {
         if(T[R].sum == T[R].rt - T[R].lt + )
         {
             return rt - lt + ;
         }
         if(lt == T[R].lt && rt == T[R].rt)
         {
             return  T[R].sum;  //返回操作处
         }
         int mid = (T[R].lt+T[R].rt)>>;
         ll ans = ;
         if(rt > mid && lt <= mid)
             ans = query(R<<,lt,mid) + query(R<<|,mid+,rt);
         else if(lt > mid)
             ans += query(R<<|,lt,rt);
         else
             ans += query(R<<,lt,rt);
         return ans;
     }
 };
 Segment_T Tr;
 int main(void)
 {
     int i,cmd,x,y,n,q,cnt= ;
     while(scanf("%d",&n) != -)
     {
         for(i = ; i <= n; i++)
             scanf("%lld",&Tr.num[i]);
         Tr.build(,,n);
         scanf("%d",&q);
         printf("Case #%d:\n",cnt++);
         for(i = ; i <q; i++)
         {
             scanf("%d %d %d",&cmd,&x,&y);
             if(y < x)
                 swap(x,y);
             if(cmd)
                 printf("%lld\n",Tr.query(,x,y));
             else
                 Tr.update(,x,y);
         }
         printf("\n");
     }
     return ;
 }