HDOJ 5063 Operation the Sequence

注意到查询次数不超过50次，那么可以从查询位置逆回去操作，就可以发现它在最初序列的位置，再逆回去即可求得当前查询的值，对于一组数据复杂度约为O(50*n)。

Operation the Sequence

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 463    Accepted Submission(s): 187

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

 

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;typedef long long int LL;const LL MOD=1000000007LL;int n,m;char op[10];int p[110000],np;LL quickpow(LL p,LL x){    LL ret=1;    LL e=p;    while(x)    {        if(x%2LL) ret=(ret*e)%MOD;        e=(e*e)%MOD;        x/=2LL;    }    return ret%MOD;}LL query(int x){    LL c=1;int pos=x;    int half=(n+1)/2;    for(int i=np-1;i>=0;i--)    {        if(p[i]==3) c=(c+c)%(MOD-1);        else if(p[i]==2) pos=n-pos+1;        else if(p[i]==1)        {            if(pos<=half) pos=2*(pos-1)+1;            else pos=(pos-half)*2;        }    }    return quickpow((LL)pos,c);}int main(){    int T_T;    scanf("%d",&T_T);    while(T_T--)    {        np=0;        scanf("%d%d",&n,&m);        for(int i=0;i<m;i++)        {            int x;            scanf("%s%d",op,&x);            if(op[0]=='O') p[np++]=x;            else if(op[0]=='Q') printf("%I64d\n",query(x));        }    }    return 0;}