P3368 【模板】树状数组 2（区间修改 单点查询）

 https://www.luogu.org/problemnew/show/P3368

//区间修改 单点查询

#include

#include

#include

#include

 

using namespace std;

 

#define lowbit(x) (x&(-x)) //计算2^k 

 

const int Max = 500000+5;

int c[Max];//c[i] = a[i – 2^k + 1] + … + a[i]，k为i在二进制下末尾0的个数, a为猿给定数组（即初始化时给的值） 

 
//单点修改，仅需修改c[i]，求区间和直接用c[i]就可以了 
void Update(int x, int y, int MAX)

{

	while(x <= MAX){

		c[x] += y;

		x += lowbit(x);

	}

}

//区间修改 （部分样例会发生	TLE） 
/*void UpdateSQ(int x, int y, int k, int MAX)
{

	while(x <= MAX){

		c[x] += k;

		x += lowbit(x);
	}
	
	while(y <= MAX){
		
		c[y] -= k;

		y += lowbit(y);
	}
	

}
 */
//a[i-lowbit(i)]+......a[i] +.....a[i-lowbit(i)-lowbit(i-lowbit(i))]+.....a[i-lowbit(i)]+.....a[0]查询 
int Sum(int x)

{

	int sum = 0;

	while(x > 0){

		sum += c[x];

		x -= lowbit(x);

	}

	return sum;

	

}

//区间查询 
void SumSQ(int y, int x)

{

	int sum = 0;

	

	while(y > 0){

		sum += c[y];

		y -= lowbit(y);

	}

	

	while(x > 0){

		sum -= c[x];

		x -= lowbit(x);

	}

	

//	cout << sum << endl;

 

	printf("%d\n", sum);	

}

int main()

{	

	int T, cnt = 1;

	string s;

	

	//cin >> T;

//	scanf("%d", &T);

	

	//while(T--)
	{

		int N, m;

		

	//	cin >> N;

		scanf("%d%d", &N, &m);

			

		memset(c, 0, sizeof(c));

		
		int tmp = 0;
		for(int i=1; i<=N; i++){

			int x;
			
			scanf("%d", &x);
//区间修改需要用到差分...... 
			Update(i, x-tmp, N);	//差分	x-tmp相当于a[i]-a[i-1] 则Sum()所求结果相当于a[i] 
			tmp = x;
		}

		

//		cout << "Case " << cnt++ << ":" << endl;
//		printf("Case %d:\n", cnt++);		

 		for(int i=0; i> x >> y;

			if(q == 1){
//区间修改
			int k;
			cin >> x >> y >> k; 
//			UpdateSQ(x, y+1, k, N);

			Update(x, k, N);
			Update(y+1, -k, N);

		}
		
		else {
			cin >> x; 
			cout << Sum(x) << endl;
		}
		
		}
	}

	return 0;

}