Secret Poems（字符串处理）

Secret Poems

The Yongzheng Emperor (13 December 1678 – 8 October 1735), was the fifth emperor of the Qing dynasty of China. He was a very hard-working ruler. He cracked down on corruption and his reign was known for being despotic, efficient, and vigorous.

Yongzheng couldn’t tolerate people saying bad words about Qing or him. So he started a movement called “words prison”. “Words prison” means literary inquisition. In the famous Zhuang Tinglong Case, more than 70 people were executed in three years because of the publication of an unauthorized history of the Ming dynasty.

In short, people under Yongzheng’s reign should be very careful if they wanted to write something. So some poets wrote poems in a very odd way that only people in their friends circle could read. This kind of poems were called secret poems.

A secret poem is a N×N matrix of characters which looks like random and meaning nothing. But if you read the characters in a certain order, you will understand it. The order is shown in figure 1 below:

            figure 1                                                           figure 2

Following the order indicated by arrows, you can get “THISISAVERYGOODPOEMITHINK”, and that can mean something.

But after some time, poets found out that some Yongzheng’s secret agent called “Mr. blood dripping” could read this kind of poems too. That was dangerous. So they introduced a new order of writing poems as shown in figure 2. And they wanted to convert the old poems written in old order as figure1 into the ones in new order. Please help them.

Input

There are no more than 10 test cases.

For each test case:

The first line is an integer N( 1 <= N <= 100), indicating that a poem is a N×N matrix which consist of capital letters.

Then N lines follow, each line is an N letters string. These N lines represent a poem in old order.

Output

For each test case, convert the poem in old order into a poem in new order.

Sample Input

5THSAD IIVOP SEOOH RGETI YMINK2ABCD4ABCDEFGHIJKLMNOP

Sample Output

THISIPOEMSDNKIAOIHTVOGYREABDCABEIKHLFNPOCMJGD

题解：

字符串处理问题，自己模拟一下，注意不同的情况就行了。

代码：

#include<stdio.h>#include<string.h>char str1[110][110],str2[11000],n,s,str3[110][110];int dfs(int x,int y,int s){    str2[s]=str1[x][y];    if(x==n-1&&y==n-1)//找到最后一个时返回；        return 0;            if((y==0&&x%2!=0&&(n%2==0&&x<n-1||n%2!=0))||(y==n-1&&n%2==0&&x%2!=0&&x<n-1||y==n-1&&n%2!=0&&x%2==0))dfs(x+1,y,s+1);  /*1、在第一列，行的下标为偶数时；（1）n为偶数，并且不在最后一行时；                                   （2）n为奇数时；    2、在最后一列时：（1）n为偶数，行的下标不为偶数，且不在最后一行时；                     （2）n为奇数，行的下标为偶数时；    以上情况都是（x,y）=>(x+1,y);  */                    else if((x==0&&y%2==0&&(n%2!=0&&y<n-1||n%2==0))||x==n-1&&(n%2==0&&y%2==0&&y<n-1||n%2!=0&&y%2!=0))dfs(x,y+1,s+1);/*  1、在第一行，列数为偶数时：（1）n为偶数；                               （2）n为奇数，且不为最后一列时；    2、在最后一行时：（1）n为偶数，列的下标为偶数，且不为最后一列时；                     （2）n为奇数，列的下标为奇数时；    以上情况都是（x,y)=>(x,y+1);*///除以上情况外：    else if((x+y)%2==0)//行和列的下标之和为偶数时：（x,y）=>（x-1,y+1）;        dfs(x-1,y+1,s+1);            else if((x+y)%2!=0)//行和列的下标之和为奇数时：（x,y）=>（x+1,y-1）;        dfs(x+1,y-1,s+1);}int print(int x,int y,int s){    str3[x][y]=str2[s];    if(s==n*n-1)return 0;    if(str3[x][y+1]=='1'&&(x-1<0||str3[x-1][y]!='1'))//在第一行或需要转弯时，向右；        print(x,y+1,s+1);    else if(str3[x+1][y]=='1')//向下；        print(x+1,y,s+1);    else if(str3[x][y-1]=='1')//向左；        print(x,y-1,s+1);    else if(str3[x-1][y]=='1')//向上；        print(x-1,y,s+1);}int main(){    int i,j;    while(~scanf("%d",&n))    {        for(i=0; i<n; i++)            scanf("%s",str1[i]);        dfs(0,0,0);        for(i=0; i<n; i++)            for(j=0; j<n; j++)                str3[i][j]='1';        print(0,0,0);        for(i=0; i<n; i++)            for(j=0; j<n; j++)                printf(j==n-1?"%c\n":"%c",str3[i][j]);    }}