1031. Hello World for U (20)

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  de  ll  rlowo

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:

helloworld!

Sample Output:

h   !e   dl   llowor

一开始看题目中上下空这么多还以为要有个空格呢，， 其实并不用

#include <iostream>#include <string>using namespace std;void print(int n1, int n2, int n3, string word){    int counter = 0;    char graph[n1][n2];    for(int i=0; i<n1; i++)    {        for(int j=0; j<n2; j++)            graph[i][j] = ' ';    }    for(int i=0; i<n1; i++)        graph[i][0] = word[counter++];    for(int i=1; i<n2-1; i++)        graph[n1-1][i] = word[counter++];    for(int i=n1-1; i>=0; i--)        graph[i][n2-1] = word[counter++];    for(int i=0; i<n1; i++)    {        for(int j=0; j<n2; j++)            cout << graph[i][j];        cout << endl;    }}int main(){    int n1, n2, n3;    string word;    cin >> word;    for(n2=3; n2<word.size(); n2++)    {        for(n1=0; n1<n2+1; n1++)        {            if(n1*2+n2-2 == word.size())                break;        }        if(n1 != n2+1)            break;    }    n1 = n3 = (word.size()+2-n2)/2;    print(n1, n2, n3, word);    return 0;}