hdu 4027 Can you answer these queries?

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 4315    Accepted Submission(s): 1036

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

 

Recommend

lcy

//2的63以内数最多开7-8次方就到1，然后就可以不用更新了
//所以复杂度为O（Nlog2N）
//此题数据没0，所以判断区间都为1就是 st[k]==r-l+1；
//坑爹的是i可能比j大，
#include <iostream>
#include <cmath>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 100003
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
using namespace std;
__int64 st[N<<2];
void build(int l,int r,int k)
{
    if(l==r)
     {
         scanf("%I64d",&st[k]);
         return ;
     }
    st[k]=0;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    st[k]=st[k<<1]+st[k<<1|1];
}
void update(int &L,int &R,int l,int r,int k)
{
    if(st[k]==r-l+1)
       return;
    if(l==r)
    {
        st[k]=sqrt(double(st[k]));
        return;
    }
    int m=(l+r)>>1;
    if(L<=m) update(L,R,lson);
    if(R>m)  update(L,R,rson);
    st[k]=st[k<<1]+st[k<<1|1];
}
__int64 query(int &L,int &R,int l,int r,int k)
{
    if(L<=l&&R>=r)
    {
        return st[k];
    }
    int m=(l+r)>>1;
    __int64 t1=0,t2=0;
    if(L<=m) t1=query(L,R,lson);
    if(R>m)  t2=query(L,R,rson);
    return t1+t2;
}
int main()
{
    int i,j,k=1;
    int n,q,op;
    while(scanf("%d",&n)!=EOF)
    {
        printf("Case #%d:\n",k++);
        build(1,n,1);
        scanf("%d",&q);
        while(q--)
        {
          scanf("%d%d%d",&op,&i,&j);
          if(i>j) {i^=j; j^=i; i^=j;}
          if(op)
          {
              printf("%I64d\n",query(i,j,1,n,1));
          }
          else
          {
              update(i,j,1,n,1);
          }
        }
        printf("\n");
    }
    return 0;
}