Operation the Sequence
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 607 Accepted Submission(s): 227
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
题目大意:给出四种操作,第一种是将奇数序的数提前,偶数序的数滞后,第二种是前后对称互换位置,第三种是每一位求取平方
题目分析:因为查询次数少,所以只要在每次查询,根据逆推公式得到最初没经过变化的位置,然后根据记录的次数取平方(注意,不是求n次方,别理解错)
逆推公式就是简单的数学换算,对于操作1的要分奇偶和前后位置分类讨论,具体见代码
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define MOD 1000000007
#define MAX 200007
using namespace std;
typedef long long LL;
LL a[MAX];
int p,cnt1,cnt2;
int t,n,m;
int bf1 ( int i )
{
if ( i > (n+1)/2 )
{
int temp = i*2-n-1;
if ( n&1 ) return temp;
else return temp+1;
}
else return i*2-1;
}
int bf2 ( int i )
{
return n+1-i;
}
void f3 ( )
{
p++;
}
LL ans[107];
char s[5];
int q[MAX],x;
int query ( int x , int cnt )
{
for ( int i = cnt-1 ; i >= 0 ; i-- )
if ( q[i] == 1 ) x = bf1(x);
else x = bf2(x);
return x;
}
int main ( )
{
scanf ( "%d" , &t );
while ( t-- )
{
p = cnt1 = cnt2 = 0;
scanf ( "%d%d" , &n , &m );
for ( int i = 1 ; i <= n ; i++ )
a[i] = i;
for ( int i = 1 ; i <= m ; i++ )
{
scanf ( "%s" , s );
scanf ( "%d" , &x );
if ( s[0] == 'O' )
{
if ( x == 3 ) f3( );
else q[cnt1++] = x;
}
else
{
int id = query ( x , cnt1 );
LL tx = a[id];
//cout << tx << endl;
for ( int i = 0 ; i < p ; i++ )
tx *= tx , tx %= MOD;
ans[cnt2++] = tx;
}
}
for ( int i = 0 ; i < cnt2 ; i++ )
printf ( "%I64d\n" , ans[i] );
}
}
