1031. Hello World for U (20)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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 lowoThat 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.
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
解:这一题卡的时间挺长,有几个测试用例一直过不去,主要是题目的条件n1 =
n3 = max { k| k <= n2 for
all 3 <= n2 <= N } with n1 +
n2 + n3 -
2 = N.这里没有注意。比如说hello,应该是
h o
e l l
还有N%3的情况需要特别注意下,我就是卡在这里,要保证n2>=n1=n3。
贴上AC代码:
#include<iostream> #include<cstdlib> #include<cstdio> using namespace std; //字符串长度为5的时候特殊考虑 int main() { string s; while(cin>>s) { // cout<<s<<endl; int len=s.length(); if(len==5) { cout<<s[0]<<" "<<s[4]<<endl; cout<<s[1]<<s[2]<<s[3]<<endl; continue; } if((len+2)%3==0) { for(int i=0;i<(len-1)/3;i++) { cout<<s[i]; for(int i=0;i<(len-4)/3;i++) cout<<' '; cout<<s[len-i-1]<<endl; } for(int i=(len-1)/3;i<(2*len+1)/3;i++) { cout<<s[i]; } cout<<endl; continue; } if((len+1)%3==0) { for(int i=0;i<(len-2)/3;i++) { cout<<s[i]; for(int i=0;i<(len-2)/3;i++) cout<<' '; cout<<s[len-i-1]<<endl; } for(int i=(len-2)/3;i<(2*len+2)/3;i++) { cout<<s[i]; } cout<<endl; continue; } if(len%3==0) { for(int i=0;i<(len)/3-1;i++) { cout<<s[i]; for(int i=0;i<(len)/3;i++) cout<<' '; cout<<s[len-i-1]<<endl; } for(int i=(len)/3-1;i<(2*len)/3+1;i++) { cout<<s[i]; } cout<<endl; continue; } } }
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