位运算 FZU - 2105 纯暴力（水，本来应该是线段树）

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

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.

看到题目的时候感觉应该是线段树，然后不是很会，然后。。。。。。。。。。。。。。

马上带来线段树版

下面是水版

这个题了你10s的时间让你算出结果，一般的oj基本上是一秒钟计算1e8,撑死1e9，然而这个题的n最坏情况是1,000,000，m最坏情况是100,000，然后t是100

最坏情况 1e15，可以说是大了好几个数量级。然后看到有几个学长超时了我就不敢写了。但是！

tmd数据水的要死，纯暴力也能过，根本不是算法题目，哎。QAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ。最后才五秒多- -！

#include
using namespace std;
int save[1000008];
int main()
{
    int t;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        getchar();
        for(int i=0; i