HDU 5063 Operation the Sequence（暴力 数学）

题目链接：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

13 5O 1O 2Q 1O 3Q 1

 

Sample Output

24

 

Source

BestCoder Round #13

PS:

把所有的操作存下来，每次把操作逆回去算一遍，求出在最初在数列中的位置，输出即可！

操作3是可以最后操作的！

代码如下：

#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;}