【樹狀數組】Mobile Phones

DescriptionSuppose 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 numberedfrom 0 to S-1. Each square contains a base station. The number of active mobile phones inside a squarecan 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 withthe row and the column of the matrix.Write a program, which receives these reports and answers queries about the current total number ofactive mobile phones in any rectangle-shaped area.InputThe input is read from standard input as integers and the answers to the queries are written to standardoutput as integers. The input is encoded as follows. Each input comes on a separate line, and consistsof 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 * 1024Cell value V at any time: 0 <= V <= 32767Update amount: -32768 <= A <= 32767No of instructions in input: 3 <= U <= 60002Maximum number of phones in the whole table: M= 2^30OutputYour program should not answer anything to lines with an instruction other than 2. If the instructionis 2, then your program is expected to answer the query by writing the answer as a single line containinga single integer to standard output.Sample Input0 41 1 2 32 0 0 2 2 1 1 1 21 1 2 -12 1 1 2 3 3Sample Output34

很裸的二維樹狀數組，注意一下0下標的處理（全部加一）。

Accode：

#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <string>#define lowbit(x) ((x) & -(x))const int maxN = 1110;int a[maxN][maxN], n;int sum(int x, int y){    if (!x || !y) return 0;    int ans = 0;    for (int x0 = x; x0; x0 -= lowbit(x0))    for (int y0 = y; y0; y0 -= lowbit(y0))        ans += a[x0][y0];    return ans;}int sum(int L, int R, int B, int T){    return sum(R, T) + sum(L, B)         - sum(R, B) - sum(L, T);}void Add(int x, int y, int v){    if (!v) return;    for (int x0 = x; x0 < n + 1; x0 += lowbit(x0))    for (int y0 = y; y0 < n + 1; y0 += lowbit(y0))        a[x0][y0] += v;    return;}inline int getint(){    int res = 0; char tmp; bool sgn = 1;    do tmp = getchar();    while (!isdigit(tmp) && tmp - '-');    if (tmp == '-') {sgn = 0; tmp = getchar();}    do res = (res << 3) + (res << 1) + tmp - '0';    while (isdigit(tmp = getchar()));    return sgn ? res : -res;}int main(){    freopen("Mobile_Phones.in", "r", stdin);    freopen("Mobile_Phones.out", "w", stdout);    getint(); n = getint();    for (int x, y, L, R, B, T, v;;)    switch (getint())    {    case 3: return 0;    case 1:        x = getint() + 1, y = getint() + 1;        v = getint(); Add(x, y, v);        break;    case 2:        L = getint(), B = getint();        R = getint() + 1, T = getint() + 1;        printf("%d\n", sum(L, R, B, T));        break;    }    return 0;}