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2013
11-12

POJ 3468 A Simple Problem with Integers [解题报告] Java

A Simple Problem with Integers

问题描述 :

You have N integers, A1, A2, … , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

输入:

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, … , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, … , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, … , Ab.

输出:

You need to answer all Q commands in order. One answer in a line.

样例输入:

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

样例输出:

4
55
9
15

温馨提示:

The sums may exceed the range of 32-bit integers.

解题代码:

//* @author
import java.util.*;
import java.io.*;

public class Main 
{
 public static void main(String []args)
 {
  SegmentTree segmentTree=new SegmentTree();
 }
};
class SegmentTree
{
 public SegmentTree()
 {
  tot=0;
  int n,m,left,right,value;
  String str;
  Scanner input=new Scanner(System.in);
  n=input.nextInt();
  m=input.nextInt();
  for(int i=1;i<=n;i++)
   a[i]=input.nextInt();
  creatTree(1,1,n);
  for(int i=0;i< m;i++)
  {
   str=input.next();
   if(str.charAt(0)=='Q')
   {
    left=input.nextInt();
    right=input.nextInt();
    System.out.println(query(1,left,right,0));
   }
   else
   {
    left=input.nextInt();
    right=input.nextInt();
    value=input.nextInt();
    insert(1,left,right,value);
   }
  }
 }
 public long creatTree(int now,int left,int right)
 {
  if(now>tot)
   tot=now;
  _left[now]=left;
  _right[now]=right;
  long lSum=0,rSum=0;
  if(left< right)
  {
   lSum=creatTree(2*now,left,(left+right)/2);
   rSum=creatTree(2*now+1,(left+right)/2+1,right); 
   _sum[now]=lSum+rSum;
  }
  else
   _sum[now]=a[left];
  return _sum[now];
 }
 public void insert(int now,int left,int right,int value)
 {
  if(now>tot)
   return ;
  if(left<=_left[now]&&right>=_right[now])
  {
   _d[now]+=value;
   return ;
  }
  long lSum=0,rSum=0;
  if(left<=(_left[now]+_right[now])/2)
   insert(2*now,left,right,value);
  if(right>(_left[now]+_right[now])/2)
   insert(2*now+1,left,right,value);
  if(2*now<=tot)
   lSum=_sum[2*now]+_d[2*now]*(_right[2*now]-_left[2*now]+1);
  if(2*now+1<=tot)
   rSum=_sum[2*now+1]+_d[2*now+1]*(_right[2*now+1]-_left[2*now+1]+1);
  _sum[now]=lSum+rSum;
 }
 public long query(int now,int left,int right,long d)
 {
  if(now>tot)
   return 0;
  if(left<=_left[now]&&right>=_right[now])
   return _sum[now]+(_d[now]+d)*(_right[now]-_left[now]+1);
  long lSum=0,rSum=0;
  if(left<=(_left[now]+_right[now])/2)
   lSum=query(2*now,left,right,d+_d[now]);
  if(right>(_left[now]+_right[now])/2)
   rSum=query(2*now+1,left,right,d+_d[now]);
  return lSum+rSum;
 }
 public static final int N=100005;
 public int tot;
 public int[] a=new int[N];
 public int[] _left=new int [3*N];
 public int[] _right=new int [3*N];
 public long[] _sum=new long [3*N];
 public long[] _d=new long [3*N];
}

  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮

  2. int half(int *array,int len,int key)
    {
    int l=0,r=len;
    while(l<r)
    {
    int m=(l+r)>>1;
    if(key>array )l=m+1;
    else if(key<array )r=m;
    else return m;
    }
    return -1;
    }
    这种就能避免一些Bug
    l,m,r
    左边是l,m;右边就是m+1,r;

  3. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?