fzu 2105 Digits Count ( 线段树 ) from 第三届福建省大学生程序设计竞赛

http://acm.fzu.edu.cn/problem.php?pid=2105

Problem Description

Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations:

Operation 1: AND opn L R

Here opn, L and R are integers.

For L≤i≤R, we do A[i]=A[i] AND opn (here "AND" is bitwise operation).

Operation 2: OR opn L R

Here opn, L and R are integers.

For L≤i≤R, we do A[i]=A[i] OR opn (here "OR" is bitwise operation).

Operation 3: XOR opn L R

Here opn, L and R are integers.

For L≤i≤R, we do A[i]=A[i] XOR opn (here "XOR" is bitwise operation).

Operation 4: SUM L R

We want to know the result of A[L]+A[L+1]+...+A[R].

Now can you solve this easy problem?

Input

The first line of the input contains an integer T, indicating the number of test cases. (T≤100)

Then T cases, for any case, the first line has two integers n and m (1≤n≤1,000,000, 1≤m≤100,000), indicating the number of elements in A and the number of operations.

Then one line follows n integers A[0], A[1], ..., A[n-1] (0≤A[i]<16,0≤i<n).

Then m lines, each line must be one of the 4 operations above. (0≤opn≤15)

Output

For each test case and for each "SUM" operation, please output the result with a single line.

Sample Input

1

4 4

1 2 4 7

SUM 0 2

XOR 5 0 0

OR 6 0 3

SUM 0 2

Sample Output

7

18

Hint

A = [1 2 4 7]

SUM 0 2, result=1+2+4=7;

XOR 5 0 0, A=[4 2 4 7];

OR 6 0 3, A=[6 6 6 7];

SUM 0 2, result=6+6+6=18.

Source

“高教社杯”第三届福建省大学生程序设计竞赛

【题解】：

　　　　这题思路想到 经过多次操作之后的区间应该是一个数字很多相同的区间：

　　 因为如果相邻两个数经过操作之后变成相同的数了，那么再经过覆盖了该区间的操作时，那么他们的值将同时发生改变，变成另外一个相同的值，这多次操作下去，之后将生更多的相同的数字区间，这里的相同数字就是懒惰标记更新的关键：例如:OR 6 0 3, A=[6 6 6 7];    其中6出现了3次

详见代码：

【code】：

 /**
 status:Accepted     language:Visual C++
 time:1953 ms   memory:47156KB
 code length:2686B   author:cj
 */
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #define N 1000010
 #define lson p<<1
 #define rson p<<1|1
 using namespace std;

 struct Nod
 {
     int l,r;
     int num;  //标记区间数字是否相同
 }node[N<<];

 void building(int l,int r,int p)
 {
     node[p].l = l;
     node[p].r = r;
     node[p].num = -;
     if(l==r)
     {
         scanf("%d",&node[p].num);
         return;
     }
     int mid = (l+r)>>;
     building(l,mid,lson);
     building(mid+,r,rson);
     if(node[lson].num!=-&&node[lson].num==node[rson].num)  //向上更新
     {
         node[p].num = node[lson].num;
     }
 }

 int opreate(int op,int opn,int num)  //操作函数 op：操作符 opn，num：操作数
 {
     if(op==)   return opn&num;
     if(op==)   return opn|num;
     if(op==)   return opn^num;

 }

 void update(int l,int r,int p,int opn,int op)
 {
     if(node[p].l==l&&node[p].r==r&&node[p].num>=)  //当找到区间并且区间被相同的数覆盖
     {
         node[p].num = opreate(op,opn,node[p].num);
         return;
     }
     if(node[p].num>=) //经过该区间时，向下更新
     {
         node[lson].num = node[rson].num = node[p].num;
         node[p].num = -;
     }
     int mid = (node[p].l+node[p].r)>>;
     if(r<=mid)  update(l,r,lson,opn,op);
     else if(l>mid)  update(l,r,rson,opn,op);
     else
     {
         update(l,mid,lson,opn,op);
         update(mid+,r,rson,opn,op);
     }
     if(node[lson].num!=-&&node[lson].num==node[rson].num)  //向上更新
     {
         node[p].num = node[lson].num;
     }
 }

 __int64 query(int l,int r,int p)
 {
     if(node[p].l==l&&node[p].r==r&&node[p].num>=)
     {
         return node[p].num*(node[p].r-node[p].l+);
     }
     if(node[p].num>=)  //经过该区间时，向下更新
     {
         node[lson].num = node[rson].num = node[p].num;
         node[p].num = -;
     }
     int mid = (node[p].l+node[p].r)>>;
     if(r<=mid)  return query(l,r,lson);
     else if(l>mid)  return query(l,r,rson);
     else return query(l,mid,lson)+query(mid+,r,rson);
     if(node[lson].num!=-&&node[lson].num==node[rson].num)  //向上更新
     {
         node[p].num = node[lson].num;
     }
 }

 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
        int n,m;
        scanf("%d%d",&n,&m);
        char op[];
        building(,n,);
        while(m--)
        {
             scanf("%s",op);
             int opn,a,b;
             if(op[]=='S')
             {
                 scanf("%d%d",&a,&b);
                 printf("%I64d\n",query(a+,b+,));
             }
             else
             {
                 scanf("%d%d%d",&opn,&a,&b);
                 if(op[]=='A')  update(a+,b+,,opn,);
                 else if(op[]=='O')  update(a+,b+,,opn,);
                 else if(op[]=='X')  update(a+,b+,,opn,);
             }
        }
     }
     return ;
 }