1031 Hello World for U （20 分)

2019-03-15

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 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:

1	helloworld!

Sample Output:

h   !
e   d
l   l
lowor

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

分析

将一个字符串按照u型输出。这里关键是确定n1,n2,n3的大小。注意n1=n3=max { k | k≤n2 for all 3≤n2≤N }，k是小于等于n2的使k=n1=n3的最大数，也就是n2>=n1=n3。那么，当总长度+2被3整除时，n1,n2,n3相等。若有余数，则加到n2上面。使用二维数组，初始化为空格，按u型填充即可。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include<sstream>

using namespace std;

int main() {
    string s;
    char c[30][30];
    cin>>s;
    int l=s.length();
    int n1,n2,n3;
    n1=n3=(l+2)/3;
    n2=n1+(l+2)%3;

    int k=0;
    fill(c[0],c[0]+30*30,' ');
    for(int i=0;i<n1-1;i++)
        c[i][0]=s[k++];
    for(int i=0;i<n2;i++)
        c[n1-1][i]=s[k++];
    for(int i=n1-2;i>=0;i--)
        c[i][n2-1]=s[k++];
    for(int i=0;i<n1;i++) {
        for (int j = 0; j < n2; j++)
            cout << c[i][j];
        cout<<endl;
    }
    return 0;
}