Mobile phones
| Time Limit: 5000MS |
|
Memory Limit: 65536K |
| Total Submissions: 7087 |
|
Accepted: 3030 |
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
一維樹狀數組用一維數組來存儲部分元素的和，二維樹狀數組只需用二維數組來存儲即可，獲得和，修正的函數同一維數組差別不大。
/**//*Source Code
Problem: 1195 User: y09
Memory: 4956K Time: 579MS
Language: C++ Result: Accepted
Source Code */
#include <stdio.h>
const int MAX=1200;
int c[MAX][MAX];
int n;
int LowBit(int t)
{
return t&(t^(t-1));
}
int Sum(int endx,int endy)
{
int sum=0;
int temp=endy;
while(endx>0)
{
endy=temp;//注意記錄endy的值，本人在此出錯，找半天錯誤不得
while (endy>0)
{
sum+=c[endx][endy];
endy-=LowBit(endy);
}
endx-=LowBit(endx);
}
return sum;
}
void plus(int posx,int posy,int num)
{
int temp=posy;
while (posx <=n)
{
posy=temp;
while(posy<=n)
{
c[posx][posy]+=num;
posy+=LowBit(posy);
}
posx+=LowBit(posx);
}
}
int GetSum(int l,int b,int r,int t)
{
return Sum(r,t)-Sum(r,b-1)-Sum(l-1,t)+Sum(l-1,b-1);
}
int main()
{
int I;
int x,y,a;
int l,b,r,t;
while(scanf("%d",&I))
{
switch (I)
{
case 0:
scanf("%d",&n);
break;
case 1:
scanf("%d%d%d",&x,&y,&a);
plus(x+1,y+1,a);
break;
case 2:
scanf("%d%d%d%d",&l,&b,&r,&t);
printf("%d\n",GetSum(l+1,b+1,r+1,t+1));
break;
case 3:
return 0;
}
}
return 0;
}