2012年浙大:Hello World for U
2016-03-24 19:44
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题目描述:
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 n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N.
输入:
There are multiple test cases.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.
输出:
For each test case, print the input string in the shape of U as specified in the description.
样例输入:
样例输出:
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 n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N.
输入:
There are multiple test cases.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.
输出:
For each test case, print the input string in the shape of U as specified in the description.
样例输入:
helloworld! ac.jobdu.com
样例输出:
h ! e d l l lowor a m c o . c jobdu. 题目来源:http://ac.jobdu.com/problem.php?pid=1464
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; char s[100]; int main() { while(scanf("%s",s)!=EOF)//gets WA { int len=strlen(s); int n1=max(min((len-1)/2,(len+2)/3),1); n1--; int n2=len-n1*2; for(int i=0;i<n1;i++) { printf("%c",s[i]); for(int j=0;j<n2-2;j++) { printf(" "); } printf("%c\n",s[len-i-1]); } for(int i=n1;i<len-n1;i++) { printf("%c",s[i]); } printf("\n"); } return 0; }
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