POJ-1195 Mobile phones 二维树状数组

　　前面用二维线段树写了个，代码长不说，而且效率还慢的要死！！！！！ Orz......

　　这题如果用树状数组来解的话，代码量很小而且速度很快。二维树状数组就在循环上面再加一层循环。

　　代码如下：

#include <cstdlib>
#include <cstdio>
#include <cstring>
#define LOWBIT( x ) (x) & -(x)
using namespace std;

int rec[1040][1040], N;

inline void CinInt( int &t )
{
    char f = 1, c;
    while( c = getchar(), c < '0' || c > '9' || c == '-' )
    {
        if( c == '-' )
            f = -1;
    }
    t = c - '0';
    while( c = getchar(), c >= '0' && c <= '9' )
        t = t * 10 + c -'0';
    t = t * f;
}

inline void CinStr( char *s )
{
    int p = -1;
    char c;
    while( c = getchar(), c == ' ' || c == '\n' ) ;
    s[++p] = c;
    while( c = getchar(), c != ' ' && c != '\n' )
        s[++p] = c;
    s[++p] = '\0';
}

inline void CinDou( double &d )
{
    d = 0;
    long long t = 0;
    int bit = 0;
    char c;
    while( c = getchar(), ( c < '0' || c > '9' ) && c != '.' ) ;
    if( c != '.' )
    {
        t = c - '0';
        while( c = getchar(), c >= '0' && c <= '9' && c != '.' )
            t = t * 10 + c - '0';
    }
    if( c == '.' )
    {
        while( c = getchar(), c >= '0' && c <= '9' )
        {
            d = d * 10 + c - '0';
            bit++;
        }
    }
    while( bit-- )
        d /= 10;
    d += t;
}

inline void modify( int x, int y, int add )
{
    for( int i = x; i <= N; i += LOWBIT( i ) )
    {
        for( int j = y; j <= N; j += LOWBIT( j ) )
            rec[i][j] += add;
    }
}

inline int query( int x, int y )
{
    int ans = 0;
    for( int i = x; i > 0; i -= LOWBIT( i ) )
    {
        for( int j = y; j > 0; j -= LOWBIT( j ) )
            ans += rec[i][j];
    }
    return ans;
}

int main()
{
    int op;
    while( scanf( "%d", &op ), op != 3 )
     {
        switch( op )
         {
            case 0:
            {
                scanf( "%d", &N );
                break;
            }
             case 1:
             {
                int x, y, add;
                scanf( "%d %d %d", &x, &y, &add );
            //    CinInt( x ), CinInt( y ), CinInt( add );
                modify( x + 1, y + 1, add );
                break;
            }
            case 2:
            {
                int sx, ex, sy, ey;
                scanf( "%d %d %d %d", &sx, &sy, &ex, &ey );
            //    CinInt( sx ), CinInt( sy ), CinInt( ex ), CinInt( ey );
                printf( "%d\n", query( ex + 1, ey + 1 ) - query( ex + 1, sy ) - query( sx, ey + 1 ) + query( sx, sy ) );
                break;
            }
        }    
    }
    return 0;
}

　　该题还有奇怪的现象就是加了输入外挂后时间竟然增加了。