树状数组简单讲解

一句话，树状数组有其特定的作用，其用途较窄，它能做的线段树都能做！

tree[1] = a1
tree[2] = tree[1] + a2;
tree[3] = a3;
tree[4] = tree[3] + tree[2] + a4 = a1 + a2 + a3 + a4;
tree[5] = a5;
tree[6] = tree[5] + a6 = a5 + a6;e
…

lowbit : 用于取出最低位的1

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

单点更新

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

区间查询（前缀和）

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

#include 
#include 
#include 
// 树状数组 
using namespace std;
// 原数组     树状数组 
int a[100010],c[100010];
int n,m;

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

// 求前缀和 
int getSum(int x){
	int ans = 0;
	while(x){
		ans += c[x];
		x -= lowbit(x);
	}
	return ans;
}

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

int main()
{
	while(scanf("%d",&n)!=EOF){
		memset(a,0,sizeof a);
		memset(c,0,sizeof c);
		
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
			update(i,a[i]);
		}
		
		int ans = getSum(n);
		cout << ans << endl;	 
	}
	
	
	return 0;
}

#include 
#include 
#include 
#include 

using namespace std;

typedef pair PII;

const int N = 100100;
int n;
int tree[N],b[N];   // b[n]存储离散化且排序后原始 pos  
PII a[N];   // 输入 a[i].fi 存储值，a[i].se 存储 pos 

// 取出二进制下第一个1及以后的数 
int lowbit(int x){
	return x & (-x);
}

// 更新点 
void update(int x,int val){
	while(x<=n){
		tree[x] += val;
		x += lowbit(x); 
	}
}

// 前缀和
int getSum(int x){
	int ans = 0;
	while(x){
		ans += tree[x];
		x -= lowbit(x);
	}
	return ans;
} 

int main()
{
	memset(tree,0,sizeof tree);
	scanf("%d",&n);
	
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i].first);
		a[i].second = i;
	}
	
	sort(a+1,a+n+1);
	
	int pos = 1;
	for(int i=1;i<=n;i++){
		if(a[i].first!=a[i-1].first && i!=1)	pos++;
		
		b[a[i].second] = pos;
	}
	
	// 求逆序对数量.
	int sum = 0;
	for(int i=1;i<=n;i++){
		update(b[i],1);
		sum += (i - getSum(b[i]));
	}	
	cout << sum << endl;
	
	return 0;
}

求逆序对(贪心+树状数组)
Array Optimization by Deque