树状数组

树状数组是与线段树实现的功能类似的数据结构，也是可以在log(n)的复杂度实现区间修改和区间查询。
以简洁的代码和思路占优，十分容易实现前缀和。

单点变化，区间查询

#include <cstdio>
using namespace std;
typedef long long ll;

ll c[500005];   //表示区间[i-lowbit(i)+1,i]的区间元素和 
int a[500005]; 
int n;

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

void update(int x,int k)   //向上更新，c[i+lowbit(i)]表示的区间一定包含了c[i] 
{
	for (int i = x; i <= n; i += lowbit(i))
	{
		c[i] += k;
	}
}

ll query(int x)    //sum[x-lowbit(x)+1,x]+sum[t-lowbit(t)+1,t]( t = x-lowbit(x) )...求出sum[1,x] 
{
	ll ans = 0;
	for (int i = x; i; i -= lowbit(i))
	{
		ans += c[i];
	}
	return ans;
}

int main()
{
	int m;
	scanf("%d%d",&n,&m);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d",&a[i]);
		update(i,a[i]);     //原来的值必须更新
	}
	while( m-- )
	{
		int kind,x,y;
		scanf("%d%d%d",&kind,&x,&y);
		if( kind == 1 )
		{
			update(x,y);
		}else
		{
			printf("%d\n",query(y) - query(x-1));
		}
	}
	return 0;
}

区间变化，单点查询
若要区间变化，那么我们维护的是数列必须是差分，这样才能保证改变的时候是只改变两个边界数，a[i]差分前缀和等于a[i]。

#include <cstdio>
using namespace std;
typedef long long ll;

ll c[500005]; 
ll a[500005];
int n;

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

void update(int x,int k)
{
	for (int i = x; i <= n; i += lowbit(i))
	{
		c[i] += k;
	}
}

ll query(int x)
{
	ll ans = 0;
	for (int i = x; i; i -= lowbit(i))
	{
		ans += c[i];
	}
	return ans;
}

int main()
{
	int m;
	ll last = 0;
	scanf("%d%d",&n,&m);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d",&a[i]);
		update(i,a[i]-last);
		last = a[i];
	}
	while( m-- )
	{
		int kind;
		scanf("%d",&kind);
		if( kind == 1 )
		{
			int x,y,k;
			scanf("%d%d%d",&x,&y,&k);
			update(x,k);
			update(y+1,-k);
		}else
		{
			int x;
			scanf("%d",&x);
			printf("%lld\n",query(x));
		}
	}
	return 0;
}

区间变化，区间查询

数学证明后得：
sum(1…a[i]) = (n+1) * sum(1…d[i]) - sum(1…t[i]) t[i] = i*d[i]

#include <cstdio>
using namespace std;
typedef long long ll;

ll sum1[500005],sum2[500005];
//sum1[i]表示d[i]构成的树状数组    d[i]为差分 
//sum2[i]表示i*d[i]构成的树状数组 
ll a[500005];
int n;

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

void update(int x,int k)
{
	for (int i = x; i <= n; i += lowbit(i))
	{
		sum1[i] += k;
		sum2[i] += x * k; 
	}
}

ll query(int x)
{
	ll ans = 0;
	for (int i = x; i; i -= lowbit(i))
	{
		ans += ( x + 1 ) * sum1[i] - sum2[i];
		//sum(a[i-lowbit(i)]...a[i]) = (n+1)*sum(d[i-lowbit[i]]...i) - sum(t[i-lowbit[i]]...i);
		//t[i] = i * d[i]; 
	}
	return ans;
}

int main()
{
	int m;
	ll last = 0;
	scanf("%d%d",&n,&m);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d",&a[i]);
		update(i,a[i]-last);
		last = a[i];
	}
	while( m-- )
	{
		int kind;
		scanf("%d",&kind);
		if( kind == 1 )
		{
			int x,y,k;
			scanf("%d%d%d",&x,&y,&k);
			update(x,k);
			update(y+1,-k);
		}else
		{
			int x;
			scanf("%d",&x);
			printf("%lld\n",query(x)-query(x-1));
		}
	}
	return 0;
}