HDU 1166 敌兵布阵 (数状数组，或线段树)

题意：。。。

析：可以直接用数状数组进行模拟，也可以用线段树。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50000 + 5;
const int mod = 1e9 + 7;
const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1};
const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int sum[maxn<<1];

int lowbit(int x){ return -x & x;  }

void add(int x, int val){
    while(x <= n){
        sum[x] += val;
        x += lowbit(x);
    }
}

int query(int x){
    int ans = 0;
    while(x){
        ans += sum[x];
        x -= lowbit(x);
    }
    return ans;
}

int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d", &n);
        memset(sum, 0, sizeof sum);
        for(int i = 1; i <= n; ++i){
            scanf("%d", &m);
            add(i, m);
        }
        char s[10];
        int u, v;
        printf("Case %d:
", kase);
        while(scanf("%s", s) == 1 && s[0] != 'E'){
            if(s[0] == 'A'){
                scanf("%d %d", &u, &v);
                add(u, v);
            }
            else if(s[0] == 'S'){
                scanf("%d %d", &u, &v);
                add(u, -v);
            }
            else{
                scanf("%d %d", &u, &v);
                printf("%d
", query(v)-query(u-1));
            }
        }
    }
    return 0;
}