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第一次写CCPC的题感觉很难，还是自己太菜了呀！！！
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 31 2 3 4 5343
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
#include#include 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;}
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
3123
Sample Output
CCPCCCCCPCPCPPCCCPPCCCCCCCCCPCPCPCPCPPCCPPCCCPPCCPPCPPPPCCCCCPCPPCPCCCPPPPCCPCCPCPPC
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。
#include#include #include 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;}
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