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1、

2、题目大意

给定4种操作，分别是

操作0的意思是创建一个S*S的空矩阵

操作1的意思是将矩阵中的（X,Y）位值的数字加上A

操作2的意思是求出左上角坐标是（L,B），右下角坐标是(R,T),求这一个矩形内的所有点的值的和

操作3是样例结束符

3、题目分析

乍一看题目好像挺简单的，不就是操作一个矩阵嘛，从头遍历查询求和就行，问题就在这，矩阵很大，查询次数多，而且这个矩阵还是在不断变化的，所以简单的二维数组时无法实现的，这道题目用到的是二维的树状数组，头不难，但是作为第一道二维的题目还是不简单的

对于最终的结果ll ans=getSum(x2,y2)-getSum(x1-1,y2)-getSum(x2,y1-1)+getSum(x1-1,y1-1);

只要画个矩形就能很容易的看出来

4、题目：

Description

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

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

5、AC代码：

#include
   
    #include
    
     #include
     
      using namespace std;#define N 1030#define ll long longint c[N][N];int n;int lowbit(int i){    return i&(-i);}void update(int x,int y,int a){    for(int i=x;i<=n;i+=lowbit(i))    {        for(int j=y;j<=n;j+=lowbit(j))        {            c[i][j]+=a;        }    }}ll getSum(int x,int y){    ll 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;}int main(){    int op,x,y,a,x1,y1,x2,y2;    while(scanf("%d",&op)!=EOF)    {        if(op==3)        break;        if(op==0)        {            scanf("%d",&n);            for(int i=1;i<=n;i++)            {                for(int j=1;j<=n;j++)                {                    c[i][j]=0;                }            }        }        else if(op==1)        {            scanf("%d%d%d",&x,&y,&a);            //下标从0开始，首先都后移一位            x++;            y++;            update(x,y,a);        }        else if(op==2)        {            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);            x1++;            y1++;            x2++;            y2++;            //printf("*%lld %lld %lld %lld\n",getSum(x2,y2),getSum(x1-1,y2),getSum(x2,y1-1),getSum(x1-1,y1-1));            ll ans=getSum(x2,y2)-getSum(x1-1,y2)-getSum(x2,y1-1)+getSum(x1-1,y1-1);            printf("%lld\n",ans);        }    }    return 0;}
     
    
   

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