Operation the Sequence（HDU 5063）

Operation the Sequence

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 440 Accepted Submission(s): 181

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 integer 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

13 5O 1O 2Q 1O 3Q 1

Sample Output

24

Source

BestCoder Round #13

题意：

给你一些操作和一些询问，每次询问都要你给出相应结果。

分析：

如果按照一个一个操作后询问时在给出答案肯定会超时，所以可以换种思维：每次询问时我再往回操作找到解（我称作懒人思维==）。由于询问次数只有50，故时间复杂度为O(50*m);故应该不会超时。

#include<stdio.h>#define N 100010#define mod 1000000007int n, notes[N], a[N];//找出最初的位置int solve(int oper, int position){    for(int i=oper-1; i>=1; i--)    {        if(notes[i] == 1)        {            if(position <= (n+1)/2) position = 2*position-1;            else position = 2*position-n-n%2;        }        else position = n-position+1;    }    return position;}int main(){    int T, m, position; //询问的位置    char s[1];    scanf("%d", &T);    while(T--)    {        scanf("%d%d", &n,&m);        int oper = 1, times = 0;        while(m--)        {            scanf("%s%d", s,&position);            if(s[0] == 'O') //记录操作            {                if(position == 3) times++;                else notes[oper++] = position;             }            else            {                __int64 ans = solve(oper, position);                for(int i=1; i<=times; i++)                    ans = ans*ans%mod;                printf("%I64d\n", ans);            }        }    }    return 0;}