2078 Problem H Secret Message

题目描述

Jack and Jill developed a special encryption method, so they can enjoy conversations without worrrying about eavesdroppers. Here is how: let L be the length of the original message, and M be the smallest square number greater than or equal to L. Add (M − L) asterisks to the message, giving a padded message with length M. Use the padded message to fill a table of size K × K, where K2= M. Fill the table in row-major order (top to bottom row, left to right column in each row). Rotate the table 90 degrees clockwise. The encrypted message comes from reading the message in row-major order from the rotated table, omitting any asterisks.

For example, given the original message ‘iloveyouJack’, the message length is L = 12. Thus the padded message is ‘iloveyouJack****’, with length M = 16. Below are the two tables before and after rotation.

Then we read the secret message as ‘Jeiaylcookuv’.

输入

The first line of input is the number of original messages, 1 ≤ N ≤ 100. The following N lines each have a message to encrypt. Each message contains only characters a–z (lower and upper case), and has length 1 ≤ L ≤ 10 000.

输出

For each original message, output the secret message.

样例输入

2iloveyoutooJillTheContestisOver

样例输出

iteiloylloooJuvOsoTvtnheiterseC
解题心得：
　　题意是输入字符串（长度L），然后将每个字符从上到下，从左到右挨着放到一个k*k的方格里，剩余的用*补上，然后将方格顺时针旋转一下（90度），然后在按照从上到下，从左到右的顺序输出（*省略掉）。K值：先把L开方，然后向上取整再加1，然后平方，即得到K的值。
　　我在做的时候忘记了把X清零，结果又是浪费了时间。。。

代码：
#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;int main(){    int n;    int x=0;    int size=0;    double k=0;    int k1=0;    int s;    char a[10000];    char c[100][100];    scanf("%d",&n);    for(int i=0;i<n;i++){        cin>>a;        size=strlen(a);        k=sqrt(size);        k1=ceil(k);        s=k1*k1;        if(size!=s){            int add=s-size;            for(int i1=0;i1<add;i1++){                a[size+i1]='*';            }            a[size+add]='\n';        }         for(int j=0;j<k1;j++){                for(int j1=0;j1<k1;j1++){                    c[j][j1]=a[x++];                }            }            x=0; //做的时候被我落掉了，结果好久才找到错误！            for(int r=0;r<k1;r++){                for(int p=k1-1;p>=0;p--){                    if(c[p][r]!='*'){                        printf("%c",c[p][r]);                    }                }            }            printf("\n");    }    return 0;}