hdu5063——Operation the Sequence

Operation the Sequence

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 429    Accepted Submission(s): 180

Problem Description

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

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

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关于这道题，我一开始是顺着给的操作推出位置的，后来发现这样做正好相反，反过来就对了

可以这么想：假设现在在位置p，
1）p是奇数，那么下一次在位置 p 前面的只有位置是奇数而且比p小的那些位置了
所以位置就是 (p+1)/2;反过来，上一次的位置p和这一次的位置p'的关系就是p = 2*p' - 1;
2) p是偶数，那么下一次p的位置就是(n+1)/ 2 + p / 2,反过来，上一次的位置p和这一次的位置p'的关系就是p = 2*p' - n - 1;显然此时p>=(n+1)/2;
其他操作就简单了

#include<set>
#include<list>
#include<queue>
#include<stack>
#include<vector>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>


using namespace std;
const int MOD = 1000000007;

int cnt;
int n, m;

int op[100010];

void solve(int t)
{
    int y = 0;
    int p = t;
    for(int i = cnt - 1; i >= 0; i--)
    {
        if(op[i] == 1)
        {
            if(p <= ((n + 1) / 2))
            {
             //   printf("cur_p == %d\n", p);
                p = p * 2 - 1;
               // printf("next_p == %d\n", p);
            }
            else
            {
                p = (p - ((n + 1) / 2)) * 2;
             //   printf("%d\n", p);
            }
        }
        else if(op[i] == 2)
        {
            p = n - p + 1;
        }
        else if(op[i] == 3)
            y++;
    }
    __int64 ans = p;
    for(int i = 1; i <= y; i++)
    {
        ans = ans * ans % MOD; 
    }
    printf("%I64d\n", ans);
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        cnt = 0;
        char C[3];
        int x;
        for(int i = 0; i < m; i++)
        {
            scanf("%s%d", C, &x);
            if(C[0] == 'O')
            {
                op[cnt++] = x;
            }
            else
            {
                solve(x);
            }
        }
    }
    return 0;
}