2019中國大學生程序設計競賽（CCPC） - 網絡選拔
1001 ^&^（HDU6702）
Problem Description
Bit operation is a common computing method in computer science ,Now we have two positive integers A and B ,Please find a positive integer C that minimize the value of the formula (A xor C) & (B xor C) .Sometimes we can find a lot of C to do this ,So you need to find the smallest C that meets the criteria .
For example ,Let's say A is equal to 5 and B is equal to 3 ,we can choose C=1,3.... ,so the answer we're looking for C is equal to 1.
If the value of the expression is 0 when C=0, please print 1.
input
The input file contains T test samples.(1<=T<=100)
The first line of input file is an integer T.
Then the T lines contains 2 positive integers, A and B, (1≤A,B<232)
Output
For each test case,you should output the answer and a line for each answer.
Sample Input
1 3 5
Sample Output
1
題意分析
找出最小的值c使（a^c）&（b^c）最大。 （位運算的水題，簽到）
#include <iostream>
#include <algorithm>
#include <string>
#include <math.h>
#include <string.h>
#include <stack>
#include <map>
using namespace std;
#define ll long long
#define fo(i, a, b) for (int i = a; i < b; i++)
#define foo(i, a, b) for (int i = a; i > b; i--)
#define mst(a, b) memset(a, b, sizeof(a))
const int maxn = 1e5+10;
int main()
{
//freopen("in.txt", "r", stdin);
ios::sync_with_stdio(false);
ll a, b;
int t;
cin >> t;
while (t--)
{
cin >> a >> b;
if ((a & b) == 0)
cout << min(a, b);
else
cout << (a & b);
cout << endl;
}
return 0;
}
1006 Shuffle Card（HDU6707）
Problem Description
A deck of card consists of n cards. Each card is different, numbered from 1 to n. At first, the cards were ordered from 1 to n. We complete the shuffle process in the following way, In each operation, we will draw a card and put it in the position of the first card, and repeat this operation for m times.
Please output the order of cards after m operations.
Input
The first line of input contains two positive integers n and m.（1<=n,m<=105）
The second line of the input file has n Numbers, a sequence of 1 through n.
Next there are m rows, each of which has a positive integer si, representing the card number extracted by the i-th operation.
Output
Please output the order of cards after m operations. (There should be one space after each number.)
Sample Input
5 3
1 2 3 4 5
3
4
3
Sample Output
3 4 1 2 5
題意分析：
最開始給出一個排列，然後m次操作，把a移動到最前面，求最後的排列（stl+簡單模擬。簽到）
解題思路：
用map進行查重模擬一遍，不論某個數字改變多少次，怎麼移動。從最後一步往前推。最後一步移動的肯定是在最前面。
然後依次在後面，往結果裏存，用map檢查該數字是否已經存入結果。
（這題因爲我數組開小了，把1e5當做1w了，超時了好幾次。我自己背鍋。在時間允許的情況下，建議開動態的。避免我這樣的問題，超時的原因吃好幾次罰時纔出來。）
#include <iostream>
#include <algorithm>
#include <string>
#include <math.h>
#include <string.h>
#include <stack>
#include <map>
using namespace std;
#define ll long long
#define fo(i, a, b) for (int i = a; i < b; i++)
#define foo(i, a, b) for (int i = a; i > b; i--)
#define mst(a, b) memset(a, b, sizeof(a))
const int maxn = 100500;
int main()
{
// freopen("in.txt", "r", stdin);
ios::sync_with_stdio(false);
map<int, int> a;
int q[maxn];
int qq[maxn];
int s[maxn];
int n, m;
while (cin >> n >> m)
{
map<int, int> a;
fo(i, 0, n)
{
cin >> q[i];
a[q[i]] = 0;
}
fo(i, 0, m)
{
cin >> qq[i];
}
foo(i, m - 1, -1)
{
if (a[qq[i]] != 1)
{
cout << qq[i] << ' ';
a[qq[i]] = 1;
}
}
fo(i, 0, n)
{
if (a[q[i]] == 0)
{
cout << q[i] << ' ';
a[q[i]] = 1;
}
}
mst(q, 0);
mst(qq, 0);
mst(s, 0);
}
}
1007 Windows Of CCPC（HDU6708）
Problem Description
In recent years, CCPC has developed rapidly and gained a large number of competitors .One contestant designed a design called CCPC Windows .The 1-st order CCPC window is shown in the figure:
And the 2-nd order CCPC window is shown in the figure:We can easily find that the window of CCPC of order k is generated by taking the window of CCPC of order k−1 as C of order k, and the result of inverting C/P in the window of CCPC of order k−1 as P of order k.
And now I have an order k ,please output k-order CCPC Windows , The CCPC window of order k is a 2k∗2k matrix.
input
The input file contains T test samples.(1<=T<=10)
The first line of input file is an integer T.
Then the T lines contains a positive integers k , (1≤k≤10)
Output
For each test case,you should output the answer .
Sample Input
3
1
2
3
Sample Output
CC
PC
CCCC
PCPC
PPCC
CPPC
CCCCCCCC
PCPCPCPC
PPCCPPCC
CPPCCPPC
PPPPCCCC
CPCPPCPC
CCPPPPCC
PCCPCPPC
題意分析：
規則 ：最開始是4個字符左下角那個和其餘3個不一樣，用最初的可以拼成第2個，把第2個分成4部分，左下角和第一個相反，也就是P變爲C，C變爲P，其餘相同。
思路：
按照上面的規則，看k的數據範圍不是很大，就用打表的方式，因爲每一個k的表都是建立在k-1的情況下。（題目雖然不難，但是理清怎麼打好這個表，不容易，心累）
#include <iostream>
#include <algorithm>
#include <string>
#include <math.h>
#include <string.h>
using namespace std;
#define ll long long
#define fo(i, a, b) for (int i = a; i < b; i++)
#define foo(i, a, b) for (int i = a; i > b; i++)
#define mst(a, b) memset(a, b, sizeof(a))
const int maxn = 1050;
char ch[maxn][maxn];
int main()
{
//freopen("in.txt", "r", stdin);
ios::sync_with_stdio(false);
mst(ch,0);
ch[0][0] = ch[0][1] = ch[1][1] = 'C';
ch[1][0] = 'P';
ll cnt = 1;
fo(k, 1, 10)
{
cnt = pow(2, k);
// cout << cnt << endl;
fo(i, cnt, 2 * cnt)//右下
fo(j, cnt, 2 * cnt)
ch[i][j] = ch[i - cnt][j - cnt];
fo(i, 0, cnt)//右上
fo(j, cnt, 2 * cnt)
ch[i][j] = ch[i][j - cnt];
fo(i, cnt, 2 * cnt)//左下倒置
fo(j, 0, cnt)
{
if (ch[i - cnt][j] == 'C')
ch[i][j] = 'P';
else
ch[i][j] = 'C';
}
}
int t;
cin >> t;
while (t--)
{
int k;
cin >> k;
int ans = pow(2, k);
fo(i, 0, ans)
{
fo(j, 0, ans)
cout << ch[i][j];
cout << endl;
}
}
}
今日體驗：
還是自己太菜，只會做這些簽到題，還花了很長時間。不能快速的理清思路，總是邊寫邊想，這樣不好。題感也越來越差了，看來還是最近沒刷題的原因啊。多刷題，提升題感。總的來說，今天打的很自卑。