hdoj_Operation the Sequence_5063

Operation the Sequence

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 261    Accepted Submission(s): 108

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

 

查询时，逆操作回去

#include <iostream>
#include <cstdio>
using namespace std;

int q[100005];

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        int st = 0, ed = 0;
        int n, m;
        scanf("%d%d", &n, &m);
        getchar();
        while (m--)
        {
            char ch;
            int temp;
            scanf("%c", &ch);
            scanf("%d", &temp);
            if (ch != 'Q')
            {
                q[ed++] = temp;
            }
            else 
            {
                int len = 0;
                int mid = (n + 1) / 2;
                for (int i = ed - 1; i >= st; i--)
                {
                    if (q[i] == 1)
                    {
                        if (temp > mid)
                            temp = (temp - mid) * 2;
                        else 
                            temp = temp + temp - 1;
                    }
                    else if (q[i] == 2)
                        temp = n + 1 - temp;
                    else if (q[i] == 3)
                    {
                        len++;
                    }
                }
                long long res = temp;
                while (len--)
                    res = res * res % 1000000007; 
                printf("%lld\n", res);
            }
            getchar();
        }
    }
    return 0;
}