HDU 1166-敌兵布阵

HDU 1166-敌兵布阵

题意：

给一个数组，有查询、增加、减少三种操作 对于每次询问 输出从i到j所有元素的和

思路:树状数组裸题

特别的 对于减少操作 只需向x位置更新-y即可

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define INF 0x3f3f3f3f
#define mod 1000000007
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define endl '\n'
#define lowbit(e) ((e)&-(e))
#define mem(n,a) memset(n,a,sizeof(n))
#define rep(i,be,en) for(int i=be;i<=en;++i)
#define pre(i,be,en) for(int i=en;i>=be;--i)
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }

using namespace std;
typedef long long ll;
const int N = 5e4 + 10;

int bitree[N];

void update(int x, int y, int n) {                  //单点更新
	for (int i = x;i <= n;i += lowbit(i)) {
		bitree[i] += y;
	}
}

int getsum(int x) {                                 // 区间求和
	int ans = 0;
	for (int i = x;i;i -= lowbit(i)) {
		ans += bitree[i];
	}
	return ans;
}

int main() {
	int t;cin >> t;
	for (int i = 1;i <= t;++i) {
		int n;cin >> n;
		for (int i = 1;i <= n;++i)bitree[i] = 0;
		for (int j = 1;j <= n;++j) {
			int val;scanf("%d", &val);
			update(j, val, n);
		}

		printf("Case %d:\n", i);

		while (1) {
			string s;cin >> s;
			if (s[0] == &#39;E&#39;)break;
			int x, y;scanf("%d%d", &x, &y);
			if (s[0] == &#39;Q&#39;) {
				printf("%d\n", getsum(y) - getsum(x - 1));
			}
			else if (s[0] == &#39;A&#39;)update(x, y, n);
			else update(x, -y, n);
		}
		
	}
	return 0;
}