第一次写CCPC的题感觉很难，还是自己太菜了呀！！！
1006题目：Shuffle Card
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
解题思路：
这个题正常去写的话会时间超限，就是每一次输入m中的数，前面的数就要整体后移，这样就会时间超限，所以这个题就是把n、m的数分别存入a[]、b[]两个数组中，然后再定义一个book[]数组，每一次输出之后就把数标记一下。
注意：这个题没有换行，不是多实例，而且每个数字后面都有一个空格
程序代码：
#include<stdio.h>
#include<string.h>
int a[100010],b[100100],book[200000];
int main()
{
int i,j,n,m,k;
memset(book,0,sizeof(book));//数组初始话为0
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);//把n个数存入数组中
for(i=m;i>=1;i--)
scanf("%d",&b[i]);//把m个数存入数组中
for(i=1;i<=m;i++)
{
if(book[b[i]]==0)//把b[]数组中的数输出并标记
{
printf("%d ",b[i]);
book[b[i]]=1;
}
}
for(i=1;i<=n;i++)
{
if(book[a[i]]==0)
{
printf("%d ",a[i]);//把a[]数组中的数输出并标记
book[a[i]]=1;
}
}
return 0;
}
1007题目：Windows Of CCPC
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
解题思路：
这个题有点找规律的迹象，就是先把前两排输入，然后从第三排开始，看第二排是P还是C，如果是C，就记录为CCPC，如果是P，就记录PPCP。
程序代码：
#include<stdio.h>
#include<string.h>
#include<math.h>
char e[2000][2000];
int main()
{
int i,j,k,n,m,t;
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
m=pow(2,n);
for(i=1;i<=m;i++)
e[1][i]='C';
for(i=1;i<=m;i++)
{
if(i%2==0)
e[2][i]='C';
else
e[2][i]='P';
}
t=2;
for(i=3;i<=m;i=i+2)
{
k=1;
for(j=1;j<=m;j=j+2)
{
if(e[t][k]=='C')
{
e[i][j]='C';
e[i][j+1]='C';
e[i+1][j]='P';
e[i+1][j+1]='C';
k++;
}
else
{
e[i][j]='P';
e[i][j+1]='P';
e[i+1][j]='C';
e[i+1][j+1]='P';
k++;
}
}
t++;
}
for(i=1;i<=m;i++)
{
for(j=1;j<=m;j++)
printf("%c",e[i][j]);
printf("\n");
}
}
return 0;
}
