HDU - 5036 Operation the Sequence
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
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, m(0.
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
題意：O1操作是將奇數位置的下標放到前面，偶數放到後面，O2是將序列翻轉，O3是序列平方，求每次Q下標的數是多少
思路：注意到查詢次數不超過50次，那麼可以從查詢位置逆回去操作，就可以發現它在最初序列的位置，再逆回去即可求得當前查詢的值，對於一組數據復雜度約為O(50*n)。
#include
#include
#include
#include
#include
typedef __int64 ll;
//typedef long long ll;
using namespace std;
const int maxn = 100005;
const int mod = 1000000007;
int n, m;
int O[maxn];
ll num[maxn];
int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
char str[10];
int op, cnt = 0;
int sum = 0;
for (int i = 0; i < m; i++) {
scanf("%s%d", str, &op);
if (str[0] == 'O') {
if (op == 3)
sum++;
else O[cnt++] = op;
}
else {
for (int j = cnt-1; j >= 0; j--) {
int cur = O[j];
if (cur == 1) {
if (n & 1) {
if (op <= n/2 + 1)
op = op * 2 - 1;
else op = (op - n / 2 - 1) * 2;
}
else {
if (op <= n / 2)
op = op * 2 - 1;
else op = (op - n / 2) * 2;
}
}
else if (cur == 2)
op = n - op + 1;
}
ll ans = op;
for (int i = 0; i < sum; i++)
ans = ans * ans % mod;
printf("%I64d\n", ans);
}
}
}
return 0;
}
