HDU 5063 Operation the Sequence（暴力 数学）

题目链接：

pid=5063" target="_blank">http://acm.hdu.edu.cn/showproblem.php?pid=5063

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

 

Source

BestCoder Round #13

PS:

把全部的操作存下来，每次把操作逆回去算一遍，求出在最初在数列中的位置。输出就可以！

操作3是能够最后操作的！

代码例如以下：

#include <cstdio>
#include <cstring>
const int maxn = 100017;
const int mod = 1000000007;
typedef __int64 LL;
int a[maxn], b[maxn];
int n, m;
int find_pos(int l, int p)
{
    for(int i = l; i > 0; i--)
    {
        if(b[i] == 1)
        {
            if (p > (n + 1) / 2)
                p = (p - (n + 1) / 2) * 2;
            else
                p = (p - 1) * 2 + 1;
        }
        else
            p = n-p+1;
    }
    return p;
}
int main()
{
    int t;
    char s[2];
    int p;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        int k = 0, l = 0;
        for(int i = 1; i <= m; i++)
        {
            scanf("%s%d",s,&p);
            if(s[0] == 'O')
            {
                if(p == 3)
                    k++;
                else
                    b[++l] = p;
            }
            else
            {
                LL ans = find_pos(l,p);
                for(int i = 1; i <= k; i++)
                {
                    ans = ans*ans%mod;
                }
                printf("%I64d\n",ans);
            }
        }
    }
    return 0;
}