hdu 1166 敌兵布阵 (树状数组)

http://acm.hdu.edu.cn/showproblem.php?pid=1166

相比与线段树，树状数组的编程难度确实很低，不容易出错，而且同等问题上效率较高些。 

但是好像树状数组能解决的问题线段树都可以解决，而线段树能解决的它却不一定能解决。

code:

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <iomanip>
#include <bitset>
#include <list>
#include <ctime>
using namespace std ;

#define SET(arr, what)  memset(arr, what, sizeof(arr))
#define FF(i, a)        for(i=0; i<a; i++)
#define SD(a)           scanf("%d", &a)
#define SSD(a, b)       scanf("%d%d", &a, &b)
#define SF(a)           scanf("%lf", &a)
#define SS(a)           scanf("%s", a)
#define SLD(a)          scanf("%lld", &a)
#define PF(a)           printf("%d\n", a)
#define PPF(a, b)       printf("%d %d\n", a, b)
#define SZ(arr)         (int)a.size()
#define SWAP(a,b)       a=a xor b;b= a xor b;a=a xor b;
#define read            freopen("in.txt", "r", stdin)
#define write            freopen("out.txt", "w", stdout)
#define MAX             1<<30
#define ESP             1e-5
#define lson            l, m, rt<<1
#define rson            m+1, r, rt<<1|1
template<class T> inline T sqr(T a){return a*a;}
template<class T> inline void AMin(T &a,T b){if(a==-1||a>b)a=b;}
template<class T> inline void AMax(T &a,T b){if(a<b)a=b;}
template<class T> inline T Min(T a,T b){return a>b?b:a;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
const int MAXN = 50010 ;
int n, c[MAXN] ;
int lowbit(int x){
    return x & (-x) ;
}
void update(int pos, int val){
    while(pos<=n){
        c[pos] += val ;
        pos += lowbit(pos) ;
    }
}
int query(int pos){
    int s = 0 ;
    while(pos>0){
        s += c[pos] ;
        pos -= lowbit(pos) ;
    }
    return s ;
}
int main(){
    char str[10] ;
    int t, i, j, T, x, v ;
    SD(t) ;
    FF(T, t){
        printf("Case %d:\n", T+1) ;
        SET(c, 0) ;
        SD(n) ;
        for(i=1; i<=n; i++){
            SD(x) ;
            update(i, x) ;
        }
        while(1){
            SS(str) ;
            if(str[0]=='E') break ;
            SSD(x, v) ;
            if(str[0]=='A')
                update(x, v) ;
            else if(str[0]=='S')
                update(x, -v) ;
            else if(str[0]=='Q')
                PF(query(v)-query(x-1)) ;
        }
    }
    return 0 ;
}