## 【数据结构】动态单链表

#include <iostream>
#include <string>
#include <cstring>
#include <queue>
using namespace std;
template<class T>
struct Node
{
T data;
Node<T> *next_pos;
};

template<class T>
{
public:
T Get(int i);
int Locate(T x);
void Insert(T x,int i);
T Delete(int i);
void PrintList();
int length();
private:
Node<T> *first,*p,*s;
};

template<class T>
{
first = new Node <T>;
first->next_pos = NULL;
}

template<class T>
{
first = new Node<T>;
first->next_pos = NULL;
for (int i = 0; i < n; i++)
{
p = new Node<T>;
p->data = arr[i];
p->next_pos = first->next_pos;
first->next_pos = p;
}
/*尾插法
first = new Node<T>;
first->next_pos = NULL;
p = first;
for (int i = 1; i <= n; i++)
{
s = new Node<T>;
s->data = arr[i];
s->next_pos = NULL;
p->next_pos = s;
p = p->next_pos;
}*/
}

template<class T>
{
p = first->next_pos;
for (int i = 0; i < n - 1 && p != NULL; i++)
{
p = p->next_pos;
}
return p == NULL ? -1 : p->data;
}

template<class T>
{
p = first->next_pos;
int count = 1;
while (p!=NULL)
{
if (p->data == x)
{
return count;
}
}
return p == NULL ? -1 : p->data;
}

template<class T>
{
p = first->next_pos;
int count = 1;
while (p!=NULL && count <= i-1)
{
p = p->next_pos;
count++;
}
if (p == NULL) throw "error";
else {
s = new Node<T>;
s->data = x;
s->next_pos = p->next_pos;
p->next_pos = s;
}
}

template<class T>
{
p = first->next_pos;
int count = 0;
while (p != NULL && count < i - 1)
{
p = p->next_pos;
count++;
}
if (p == NULL||p->next_pos == NULL) throw "error";
else {
s = p->next_pos;
p->next_pos = p->next_pos->next_pos;
T ans=s->data;
delete s;
return ans;
}

}

template<class T>
{
p=first->next_pos;
while (p != NULL)
{
cout << p->data << " ";
p = p->next_pos;
}
cout << endl;
}

template<class T>
{
p = first->next_pos;
long long int count = 0;
while (p != NULL)
{
p = p->next_pos; count++;
}
return count;
}
int main()
{
int a[9999];
for (int i = 0; i < 100; i++)
a[i] = i;
ab.PrintList();
ab.Insert(100, 2);
ab.PrintList();
cout << ab.length() << endl;
cout << ab.Get(2) << endl;
ab.Delete(2);
ab.PrintList();
cout << ab.length() << endl;
cout << ab.Get(2) << endl;
ab.PrintList();
}

## 【树状数组】Curious Robin Hood

Robin Hood likes to loot rich people since he helps the poor people with this money. Instead of keeping all the money together he does another trick. He keeps n sacks where he keeps this money. The sacks are numbered from 0 to n-1.

Now each time he can he can do one of the three tasks.

1)                  Give all the money of the ith sack to the poor, leaving the sack empty.

2)                  Add new amount (given in input) in the ith sack.

3)                  Find the total amount of money from ith sack to jth sack.

Since he is not a programmer, he seeks your help.