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Operation the Sequence
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 158 Accepted Submission(s): 74
Problem Description
You have an array consisting of n integers:
a1=1,a2=2,a3=3,…,an=n
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];
}
a1=1,a2=2,a3=3,…,an=n
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)
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.
T(1≤T≤20)
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
官方题解
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)。
ps：记两个操作数组a和c，数组a存的是奇偶排序的上一个元素的位置，数组c存的是逆置操作的上一个元素的位置，这样就能够逆回去操作了。
代码：
//218ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=100000+1000;
int a[maxn];//奇偶排序操作
int q[maxn];//存储操作类型，1是奇偶排序。2是逆置
int c[maxn];//逆置
const int mod=1000000007;
int solve(int cur,int x)//找到在刚開始的位置
{
int ans=x;
for(int i=cur-1;i>=0;i--)
{
if(q[i]==1)
{
ans=a[ans];
}
else
{
ans=c[ans];
}
}
return ans;
}
int main()
{
int t,n,m;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int index=1;
for(int i=1; i<=n; i +=2)
a[index++]=i;
for(int i=2; i<=n; i +=2)
a[index++]=i;
for(int i=1;i<=n;i++)
c[i]=n+1-i;
char s[10];
int p;
int cur=0;
int cou=0;
for(int i=0;i<m;i++)
{
scanf("%s%d",s,&p);
if(s[0]=='O')
{
if(p==3)
cou++;
else
q[cur++]=p;
}
else
{
long long ans=solve(cur,p);
for(int i=0;i<cou;i++)
{
ans=ans*ans%mod;
}
printf("%I64d\n",ans);
}
}
}
return 0;
}
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