HDOJ 5063 Operation the Sequence （优化后的暴力）

Operation the Sequence

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1000    Accepted Submission(s): 341

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 integerT(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

注意到“this operator will have at most 50”（Q），所以可优化下后用暴力，复杂度由O（m*n）降至O（50*m）。

#include <bits/stdc++.h>using namespace std;#define mst(a,b) memset((a),(b),sizeof(a))#define f(i,a,b) for(int i=(a);i<=(b);++i)const int maxn =100005;const int mod = 1e9+7;const int INF = 1e9;#define ll long long#define rush() int T;scanf("%d",&T);while(T--)int m,n,op[maxn]; char s[2];int cnt;ll pos;ll query()  //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{    int b=0;    for(int i=cnt-1; i>=0; --i)    {        if(op[i]==1)        {            if(pos<=(n+1)/2)            {                pos=2*pos-1;            }            else            {                pos=2*(pos-(n+1)/2);            }        }        else if(op[i]==2)        {            pos=n-pos+1;        }        else if(op[i]==3)        {            b++;        }    }    while(b--)    {        pos=pos*pos%mod;    }    return pos;}int main(){    rush()    {        scanf("%d%d",&n,&m);        cnt=0;        while(m--)        {            scanf("%s%I64d",s,&pos);            if(s[0]=='O')            {                op[cnt++]=pos;            }            else if(s[0]=='Q')            {                printf("%I64d\n",query());            }        }    }    return 0;}