poj 1195 Mobile phones

Mobile phones

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 10930		Accepted: 5002

Description

Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contains a base station. The number of active mobile phones inside a square can change because a phone is moved from a square to another or a phone is switched on or off. At times, each base station reports the change in the number of active phones to the main base station along with the row and the column of the matrix.  

Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area.  

Input

The input is read from standard input as integers and the answers to the queries are written to standard output as integers. The input is encoded as follows. Each input comes on a separate line, and consists of one instruction integer and a number of parameter integers according to the following table.  

The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3.  

Table size: 1 * 1 <= S * S <= 1024 * 1024  
Cell value V at any time: 0 <= V <= 32767  
Update amount: -32768 <= A <= 32767  
No of instructions in input: 3 <= U <= 60002  
Maximum number of phones in the whole table: M= 2^30  

Output

Your program should not answer anything to lines with an instruction other than 2. If the instruction is 2, then your program is expected to answer the query by writing the answer as a single line containing a single integer to standard output.

Sample Input

0 4

1 1 2 3

2 0 0 2 2 

1 1 1 2

1 1 2 -1

2 1 1 2 3 

3

Sample Output

3

4

Source

IOI 2001

题意：二维树状数组p[i][j]代表的i-lowbit(i)+1 to i ,j-lowbit(j)+1 to j 这个矩阵的和，insert和getsum函数做了一下改变，用两个for来实现插入和求和。又忘记 树状数组 不能为 0，调试了一会，再就是 对求某个区间的和，还是不是很熟悉wa了几次。如求（x1,y1)到（x2,y2)的和，画画图像可以很快得到公式。

getsum(x2,y2)+getsum(x1-1,y1-1)-getsum(x2,y1-1)-getsum(x1-1,y2);

AC代码：

#include <stdio.h>
int n;
int c[1500][1500];
int lowbit( int x)
{
 return x&(-x);
}
void updata( int x,int y,int a)
{
  int i,j;
  for( i=x;i<=n;i+=lowbit(i))
  {
 for( j=y;j<=n;j+=lowbit(j))
 {
   c[i][j]+=a;
    }
  }
}
int sum( int x,int y)
{
 int i,j,s=0;
 for( i=x;i>0;i-=lowbit(i))
 {
   for( j=y;j>0;j-=lowbit(j))
   {
 s+=c[i][j];
   }
 }

 return s;
}
int main( )
{
  int x1,y1,x2,y2,a;
  int index1;
 
  while(scanf("%d",&index1)!=EOF)
  {
  int i,j;
     if(index1>=3) break;
  if(index1==1)
 {
  scanf("%d%d%d",&x1,&y1,&a);
  x1++;y1++;
  updata(x1,y1,a);
 }
 if(index1==0)
 {
   scanf("%d",&n);
   n++;
   for( i=0;i<=n;i++)
   {
  for( j=0;j<=n;j++)
  {
    c[i][j]=0;
  }
   }
    }
 if(index1==2)
    {
 
   scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
   x1++;y1++;
   x2++;y2++;
   int xx=sum(x2,y2)+sum(x1-1,y1-1)-sum(x2,y1-1)-sum(x1-1,y2);
   printf("%d\n",xx);
 
  
 }
  }
  return 0;
}

链接：http://poj.org/problem?id=1195