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 n1characters, 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
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<map> #define MAXSIZE 100005 typedef long long ll; using namespace std; //*******常用基本函数******* //*******本题用到函数******* //变为U void varyU(char* s) { int i,j,n1,n2,n3; int len = strlen(s); n1 = (len + 2) / 3 ; n2 = len - 2*n1; for(i=0;i<n1-1;i++) { cout<<s[i]; for(j=0;j<n2;j++)cout<<" "; cout<<s[strlen(s)-1-i]<<endl; } for(int k=i;k<len-i;k++)cout<<s[k]; cout<<endl; } int main() { char ss[MAXSIZE]; cin>>ss; varyU(ss); //cout<<ss[2]<<endl; return 0; }