Mobile phones

Time Limit: 5000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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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

 #include <iostream>
 #include <string.h>
 #include <stdio.h>

 using namespace std;
 #define max 1025
 int c[max][max];
 int n;

 int lowbit(int x)
 {
     return x&-x;
 }

 void update(int x,int y,int val)
 {
     for(int i=x;i<=n;i+=lowbit(i))
     {
         for(int j=y;j<=n;j+=lowbit(j))
         {
             c[i][j]+=val;
         }
     }
 }

int get_sum(int x,int y)
 {
     int s=0;
     for(int i=x;i>0;i-=lowbit(i))
     {
         for(int j=y;j>0;j-=lowbit(j))
         {
             s+=c[i][j];
         }
     }
     return s;
 }

 int main()
    {
        int i,j,s,T,x,y,x1,y1,val;
        while(1)
        {
            scanf("%d",&T);
            if (T==3) break;
            if (T==0)
            {
                scanf("%d",&n);
                memset(c,0,sizeof(c));
            }
            if (T==1)
            {
                scanf("%d%d%d",&x,&y,&val);
                update(x+1,y+1,val);   ///脚标从0开始  所以要+1喽~~~~
            }
            if (T==2)
            {
                scanf("%d%d%d%d",&x,&y,&x1,&y1);  ///同理
                printf("%d\n",get_sum(x1+1,y1+1)+get_sum(x,y)-get_sum(x1+1,y)-get_sum(x,y1+1));
            }
        }
        return 0;
    }

poj 1195 单点更新 区间求和