poj 1195 单点更新 区间求和
Mobile phones
Time Limit: 5000 MS Memory Limit: 65536 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
Description
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
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
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=;
for(int i=x;i>;i-=lowbit(i))
{
for(int j=y;j>;j-=lowbit(j))
{
s+=c[i][j];
}
}
return s;
} int main()
{
int i,j,s,T,x,y,x1,y1,val;
while()
{
scanf("%d",&T);
if (T==) break;
if (T==)
{
scanf("%d",&n);
memset(c,,sizeof(c));
}
if (T==)
{
scanf("%d%d%d",&x,&y,&val);
update(x+,y+,val); ///脚标从0开始 所以要+1喽~~~~
}
if (T==)
{
scanf("%d%d%d%d",&x,&y,&x1,&y1); ///同理
printf("%d\n",get_sum(x1+,y1+)+get_sum(x,y)-get_sum(x1+,y)-get_sum(x,y1+));
}
}
return ;
}
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