Mobile phones

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 11696		Accepted: 5369

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 41 1 2 32 0 0 2 2 1 1 1 21 1 2 -12 1 1 2 3 3

Sample Output

34

Source

 

2、解题思路：题意，给定n*n矩阵，和几种在线操作，包括对某一点(x,y)值修改，查询一个矩形(x1,y1,x2,y2)的元素和。步骤，（1）树状数组的经典应用；（2）面积处理过程中，面积=Sum(R,T)-Sum(R,B-1)-Sum(L-1,T)+Sum(L-1,B-1)。

3、注意事项：注意貌似除了树状数组这种方法，其他的方法都TLE，面积处理时候，记得加上减去两次的部分Sum(L-1,B-1)。

 

#include
      
       #include
       
        using namespace std;int N,c[1050][1050];int lowbit(int x){    return x&(-x);}int Sum(int x,int y){    int sum=0;    for(int i=x;i>0;i-=lowbit(i))        for(int j=y;j>0;j-=lowbit(j))            sum+=c[i][j];    return sum;}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 main(){    int op;    int x,y,a;    int l,b,r,t;    scanf("%d%d",&op,&N);    while(scanf("%d",&op) && op!=3){        if(op==1){            scanf("%d%d%d",&x,&y,&a);            update(x+1,y+1,a);        }else if(op==2){            scanf("%d%d%d%d",&l,&b,&r,&t);            int ans=Sum(r+1,t+1)-Sum(r+1,b)-Sum(l,t+1)+Sum(l,b);            printf("%d\n",ans);        }    }    return 0;}