POJ 2039 To and Fro

Mo和Larry发明了一种加密消息的方法,通过设定列数并将字母写成矩形阵列来实现加密。加密过程涉及交替从左到右和从右到左书写每行的字符。任务是从加密的消息中恢复原始文本。

Description
Mo and Larry have devised a way of encrypting messages. They first decide secretly on the number of columns and write the message (letters only) down the columns, padding with extra random letters so as to make a rectangular array of letters. For example, if the message is “There’s no place like home on a snowy night” and there are five columns, Mo would write down
t o i o y
h p k n n
e l e a i
r a h s g
e c o n h
s e m o t
n l e w x
Note that Mo includes only letters and writes them all in lower case. In this example, Mo used the character “x” to pad the message out to make a rectangle, although he could have used any letter.
Mo then sends the message to Larry by writing the letters in each row, alternating left-to-right and right-to-left. So, the above would be encrypted as
toioynnkpheleaigshareconhtomesnlewx
Your job is to recover for Larry the original message (along with any extra padding letters) from the encrypted one.


【题目分析】
没什么好说的,模拟小水题。


【代码】

#include <iostream>
using namespace std;
int main(){
    char str[205][205],name[250];
    int n,m,i,j,k,flag;
    while(cin>>n &&n){
        cin>>name;
        j = i = m =0;
        flag = 1;
        while(name[m]){         
            if(j<n &&flag ==1){  
                str[i][j] = name[m];
                m++;j++;
            }
            else if(j==n){        
                j--;flag = 2;i++;
                str[i][j] =name[m];
                m++;j--;
            }
            else if(j>=0 &&flag ==2){  
                str[i][j]=name[m];
                m++;j--;
            }
            else if(j<0){              
                j++;flag = 1;i++;
                str[i][j]=name[m];
                m++;j++;
            }
        }
        k = i;
        for(j=0;j<n;j++){             
            for(i=0;i<=k;i++){
                cout<<str[i][j];
            }
        }
        cout<<endl;
    }
    return 0;
}
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