POJ 1195 Mobile phones

  
    

   
     //
   
     二维数状数组 需要注意的是下标从1开始 lowbit(0) = 0.. 超时陷阱，一定要谨记啊！ 还有是c++的输入输出2600ms scanf 547ms
   
     

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

   
     using
   
      
   
     namespace
   
      std;

   
     const
   
      
   
     int
   
      MAXN 
   
     =
   
      
   
     1025
   
     ;

   
     int
   
      tree[MAXN][MAXN];

   
     int
   
      p[MAXN];

   
     int
   
      n; 
   
     //
   
     坐标范围 1 ~ 1
   
     

   
     inline 
   
     int
   
      lowbit(
   
     int
   
      t){
 
   
     return
   
      t 
   
     &
   
      (
   
     -
   
     t);
}

   
     void
   
      add(
   
     int
   
      x, 
   
     int
   
      y, 
   
     int
   
      value){
 
   
     for
   
     (;x 
   
     <=
   
      n; x 
   
     +=
   
      lowbit(x)){
 
   
     for
   
     (
   
     int
   
      tempy 
   
     =
   
      y; tempy 
   
     <=
   
      n; tempy 
   
     +=
   
      p[tempy])
   
     //
   
     lowbit(tempy))
   
     

   
      tree[x][tempy] 
   
     +=
   
      value;
 }
}

   
     int
   
      sum(
   
     int
   
      x, 
   
     int
   
      y){
 
   
     int
   
      ans 
   
     =
   
      
   
     0
   
     ;;
 
   
     for
   
     (; x 
   
     >
   
      
   
     0
   
     ; x 
   
     -=
   
      lowbit(x)){
 
   
     for
   
     (
   
     int
   
      tempy 
   
     =
   
      y; tempy 
   
     >
   
      
   
     0
   
     ; tempy 
   
     -=
   
      p[tempy])
   
     //
   
     lowbit(tempy))
   
     

   
      ans 
   
     +=
   
      tree[x][tempy];
 }
 
   
     return
   
      ans;
}

   
     int
   
      main(){
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ;i 
   
     <=
   
      
   
     1024
   
     ; i
   
     ++
   
     ) 
 p[i] 
   
     =
   
      i 
   
     &
   
      (
   
     -
   
     i);
 
   
     int
   
      cmd;
 
   
     //
   
     cin >> cmd >> n;
   
     

   
      scanf(
   
     "
   
     %d%d
   
     "
   
     ,
   
     &
   
     cmd, 
   
     &
   
     n);
 
   
     while
   
     (scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     cmd) 
   
     ==
   
      
   
     1
   
     ){
   
     //
   
     cin >> cmd){
   
     

   
      
   
     switch
   
     (cmd)
 {
 
   
     case
   
      
   
     1
   
     :
 
   
     int
   
      x, y, value;
 
   
     //
   
     cin >> x >> y >> value;
   
     

   
      scanf(
   
     "
   
     %d%d%d
   
     "
   
     ,
   
     &
   
     x,
   
     &
   
     y,
   
     &
   
     value);
 add(x 
   
     +
   
      
   
     1
   
     , y 
   
     +
   
      
   
     1
   
     , value);
 
   
     break
   
     ;
 
   
     case
   
      
   
     2
   
     :
 
   
     int
   
      x1, y1, x2, y2;
 
   
     //
   
     cin >> x1 >> y1 >> x2 >> y2;
   
     

   
      scanf(
   
     "
   
     %d%d%d%d
   
     "
   
     ,
   
     &
   
     x1,
   
     &
   
     y1,
   
     &
   
     x2,
   
     &
   
     y2);
 
   
     /*
   
     cout << sum(x2 + 1, y2 + 1) + sum(x1, y1) - sum(x1, y2 + 1) - sum(x2 + 1, y1) << endl; //下标要加一 这题目做的一波三折 我怎么写成sum(x2 + 1, y2 + 1) - sum(x1, y1)了呢 ？ 我靠..脑袋抽筋啊
   
     */
   
     
 printf(
   
     "
   
     %d\n
   
     "
   
     ,sum(x2 
   
     +
   
      
   
     1
   
     , y2 
   
     +
   
      
   
     1
   
     ) 
   
     +
   
      sum(x1, y1) 
   
     -
   
      sum(x1, y2 
   
     +
   
      
   
     1
   
     ) 
   
     -
   
      sum(x2 
   
     +
   
      
   
     1
   
     , y1));
 
   
     break
   
     ;
 
   
     default
   
     :
 
   
     return
   
      
   
     0
   
     ;
 }
 }
 
   
     return
   
      
   
     0
   
     ;
}