TZOJ--4163: Encrypt（模拟）

4163: Encrypt

描述

Every evening, little Ivica sends secret messages to little Marica through e-mail. Knowing Ivica's e-letter travels unguarded through the network on its way to Marica's e-mailbox, they have decided to encrypt every message using the following algorithm:
· Suppose Ivica's message consists of N characters.
· Ivica must first find a matrix consisting of R rows and C columns such that R ≤ C and R·C = N. If there is more than one such matrix, Ivica chooses the one with the most rows.
· Ivica writes his message into the matrix in row-major order. In other words, he writes the first segment of the message into the first row, the second segment into the second row and so on.
· The message he sends to Marica is the matrix read in column-major order.
Marica has grown tired of spending her precious time deciphering Ivica's messages, so you must write a program to do it for her.

输入

The input contains the received message, a string of lowercase letters of the English alphabet (with no spaces).
The number of letters will be between 1 and 100.

输出

Output the original (decrypted) message.

样例输入

 koaski

样例输出

 kakosi

题目来源

TOJ

 

题目链接：http://tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=4163

 

题目大意：将一个字符串，转换成一个R*C的矩形字符串，要求R和C的值最接近，并且R*C=N，R<C，然后按照这个矩形输出

要求R和C最接近，那么R=C的时候是接近的，所以我们从R=sqrt（N）开始向下枚举到可以被整除为止，这样的R和C就是满足题目要求的，后面就按照题目意思模拟。

 

#include <math.h>
#include <string.h>
#include <stdio.h>
int main()
{
	char a[111];
	char b[111][111];
	while(scanf("%s",a)){
		int n=strlen(a);
		int k=sqrt(n+1);
		//从最接近的开始向下枚举K，直到N能被K整除 
		while(n%k)k--;
		
		int i,j,l=0;
		
		//按照题目意思转换成二维数组 
		for(i=0;i<n/k;i++){
			for(j=0;j<k;j++){
				b[i][j]=a[l++];
			}
		}
		//按照题目意思从前往后遍历输出 
		for(j=0;j<k;j++){
			for(i=0;i<n/k;i++){
				printf("%c",b[i][j]);
			}
		}
		puts("");
	}
}