1031 Hello World for U (20 分)
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 n
1
?
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!
結尾無空行
Sample Output:
h !
e d
l l
lowor
結尾無空行`

自己寫的，暴力求解 AC

#include<cstdio> #include<cstring> int main() {char s[100];int n1,n2,n3;scanf("%s",s);int l=strlen(s);n3=0;for(n2=3;n2<=l;n2++){for(n1=1;n1<=n2;n1++){if(2*n1+n2-2==l){if(n1>n3)n3=n1;}}}n2=l+2-2*n3;for(int i=0;i<n3;i++){if(i!=n3-1){printf("%c",s[i]);for(int j=0;j<n2-2;j++)printf(" ");printf("%c\n",s[l-i-1]);}else{for(int j=0;j<n2;j++){printf("%c",s[i+j]);}}}return 0;}

但其實 n1=(N+2)/3 可以推理出來
或者 柳神的分析：假設n = 字符串長度 + 2，因為2 * n1 + n2 = n，且要保證n2 >= n1, n1盡可能地大，分類討論：

如果n % 3 == 0，n正好被3整除，直接n1 == n2 == n3;

如果n % 3 == 1，因為n2要比n1大，所以把多出來的那1個給n2

如果n % 3 == 2, 就把多出來的那2個給n2

所以得到公式：n1 = n / 3，n2 = n / 3 + n % 3

把它們存儲到二維字符數組中，一開始初始化字符數組為空格，然后按照u型填充進去，最后輸出這個數組u。

總結

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