A1031 Hello World for U (20 point(s))

字符串按格式输出

1. 原文

原题

Given any string of N (≥5) characters, you are asked to form the characters into the shape of U. For example, helloworld can be printed as:

h  d
e  l
l  r
lowo

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with $n_2$ characters, and finally bottom-up along the vertical line with $n_3$ characters. And more, we would like U to be as squared as possible – that is, it must be satisfied that $n_1$=$n_3$=max { k | k≤$n_2$ for all 3≤$n_2$≤N } with $n_1$+$n_2$+$n_3$−2=N.

Input Specification:

Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

output Specification:

For each test case, print the input string in the shape of U as specified in the description.

Sample Input:

helloworld!

Sample output:

h   !
e   d
l   l
lowor

2. 解析

结构存入数组中 再输出

3. AC代码

#include<cstdio>
#include<cstring>
const int maxn = 100;
char str[maxn];
char matrix[maxn][maxn];
int main(){
	for (int i = 0; i < maxn; ++i)
	{
		for (int j = 0; j < maxn; ++j)
		{
			matrix[i][j]=' ';
		}
	}
	scanf("%s",str);
	int len = strlen(str)+2;
	int n1 = len/3;
	int n2 = len/3+len%3;
	int idx = 0;
	for (int i = 0; i < n1-1; ++i)
	{
		matrix[i][0]=str[idx++];
	}
	for (int i = 0; i < n2; ++i)
	{
		matrix[n1-1][i]=str[idx++];
	}
	for (int i = n1-2; i >= 0; i--)
	{
		matrix[i][n2-1]=str[idx++];
	}
	for (int i = 0; i < n1; ++i)
	{
		for (int j = 0; j < n2; ++j)
		{
			printf("%c",matrix[i][j]);
		}
		printf("\n");
	}

	return 0;
}

Way2:

直接根据图形输出

#include<cstdio>
#include<cstring>
char str[100];
int main(){
	freopen("A1031.txt","r",stdin);
	scanf("%s",str);
	int len = strlen(str)+2;
	int n1 = len/3;
	int n2 = len/3+len%3;
	len-=2;
	int idx = 0;
	for (int i = 0; i < n1-1; ++i)
	{
		printf("%c",str[idx]);
		for (int i = 1; i < n2-1; ++i)
		{
			printf(" ");
		}
		printf("%c\n",str[len-1-idx]);
		idx++;
	}
	for (int i = 0; i < n2; ++i)
	{
		printf("%c",str[idx++]);
	}
	printf("\n");
	
	return 0;
}