HDU - 5036 Operation the Sequence
2016-01-03 19:16
639 查看
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
Sample Output
Source
BestCoder Round #13
题意:O1操作是将奇数位置的下标放到前面。偶数放到后面,O2是将序列翻转,O3是序列平方,求每次Q下标的数是多少
思路: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)。
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
1 3 5 O 1 O 2 Q 1 O 3 Q 1
Sample Output
2 4
Source
BestCoder Round #13
题意:O1操作是将奇数位置的下标放到前面。偶数放到后面,O2是将序列翻转,O3是序列平方,求每次Q下标的数是多少
思路: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)。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> 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; }
相关文章推荐
- iOS UI 沙盒路径的获取及文件的简单存储
- 【LWJGL2 WIKI】【辅助库篇】Slick-Util库:第三部分-读取TrueType字体
- easyui datagrid使用(好)
- UIAlertController在8以下不支持
- UIAlertController在8以下不支持
- 熔断器模式(CircuitBreaker)
- Android开发之Buidler模式初探结合AlertDialog.Builder讲解
- Easyui-DataGrid 的增删查改
- request对象
- 解决 UITableViewCell的点击事件和手势的冲突问题
- 并发队列ConcurrentLinkedQueue和阻塞队列LinkedBlockingQueue用法
- HDFS2—SequenceFile(小文件的解决方案)
- Android 子线程更新 UI
- 167,Xcode 3D查看UI布局效果
- UI组件的呈现
- QUESTION 47 How many copies of the alert log are stored in the directory specified by the initializa
- UUID 获取设备的唯一标志
- UVA 11133 - Eigensequence DP
- iOS开发笔记--如何去掉UItableview header(footer)view黏性(sticky)
- EasyUI Tree节点拖动到指定容器