Operation the Sequence

                                                                    Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

                                                                                            Total Submission(s): 158    Accepted Submission(s): 74

Problem Description

You have an array consisting of n integers: 

a1=1,a2=2,a3=3,…,an=n. Then give you m operators, you should process all the operators in order. Each operator is one of four types:

Type1: O 1 call fun1();

Type2: O 2 call fun2();

Type3: O 3 call fun3();

Type4: Q i query current value of a[i], 

this operator will have at most 50.

Global Variables: a[1…n],b[1…n];

fun1() {

index=1;

  for(i=1; i<=n; i +=2) 

    b[index++]=a[i];

  for(i=2; i<=n; i +=2)

    b[index++]=a[i];

  for(i=1; i<=n; ++i)

    a[i]=b[i];

}

fun2() {

  L = 1;R = n;

  while(L<R) {

    Swap(a[L], a[R]); 

    ++L;--R;

  }

}

fun3() {

  for(i=1; i<=n; ++i) 

    a[i]=a[i]*a[i];

}

 

Input

The first line in the input file is an integer 

T(1≤T≤20), indicating the number of test cases.

The first line of each test case contains two integer 

n(0<n≤100000), 

m(0<m≤100000).

Then m lines follow, each line represent an operator above.

 

Output

For each test case, output the query values, the values may be so large, you just output the values mod 1000000007(1e9+7).

 

Sample Input

1 3 5 O 1 O 2 Q 1 O 3 Q 1

 

Sample Output

2 4

 

官方题解

注意到查询次数不超过50次。那么能够从查询位置逆回去操作，就能够发现它在最初序列的位置，再逆回去就可以

求得当前查询的值。对于一组数据复杂度约为O(50*n)。

ps：记两个操作数组a和c，数组a存的是奇偶排序的上一个元素的位置，数组c存的是逆置操作的上一个元素的位置，这样就能够逆回去操作了。

  代码：

//218ms#include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;const int maxn=100000+1000;int a[maxn];//奇偶排序操作int q[maxn];//存储操作类型，1是奇偶排序。2是逆置int c[maxn];//逆置const int mod=1000000007;int solve(int cur,int x)//找到在刚開始的位置{    int ans=x;    for(int i=cur-1;i>=0;i--)    {        if(q[i]==1)        {           ans=a[ans];        }        else        {            ans=c[ans];        }    }    return ans;}int main(){    int t,n,m;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);       int index=1;       for(int i=1; i<=n; i +=2)       a[index++]=i;       for(int i=2; i<=n; i +=2)        a[index++]=i;       for(int i=1;i<=n;i++)       c[i]=n+1-i;        char s[10];        int p;        int cur=0;        int cou=0;        for(int i=0;i