hdu 5063 Operation the Sequence（Bestcoder Round #13)

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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
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<m;i++)
        {
            scanf("%s%d",s,&p);
            if(s[0]=='O')
            {
                if(p==3)
                cou++;
                else
                q[cur++]=p;
            }
            else
            {
                long long ans=solve(cur,p);

                for(int i=0;i<cou;i++)
                {
                    ans=ans*ans%mod;
                }
                printf("%I64d\n",ans);
            }
        }
    }
    return 0;
}