HDOJ 5063 Operation the Sequence

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)。

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

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

 

Sample Output

2
4

 

Source

field=problem&key=BestCoder+Round+%2313&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">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;
}

版权声明：本文博客原创文章。博客，未经同意，不得转载。