1031. Hello World for U (20)-PAT甲级真题

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 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:
helloworld!

Sample Output:
h　　　　 !
e　　　　d
l 　　　　 l
　lowor

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分析：这道题就是把一个字符串编程U型输出，n1，n3就是左右两边竖着的，这两个值相等，n2就是U型底部的，要大于n1，n2。
　　既然这样的话，就先把整个字符串平均分成３部分。因为要求n2>=n1，n1还要尽可能大，也就是n%3这多出来的不能给侧面，只可能给n2，所以。
　　n1=n/3，n2=n/3+n%3

#include <cstdio>#include <string.h>using namespace std;int main() {    char c[81];    char u[30][30];    memset(u, ' ', sizeof(u));    scanf("%s", c);    int n = strlen(c) + 2;    int n1 = n / 3;    int n2 = n1 + n % 3;    int index = 0;    for(int i = 0; i < n1; i++) {        u[i][0] = c[index++];    }    for(int i = 1; i <= n2 - 2; i++) {        u[n1-1][i] = c[index++];    }    for(int i = n1 - 1; i >= 0; i--) {        u[i][n2-1] = c[index++];    }    for(int i = 0; i < n1; i++) {        for(int j = 0; j < n2; j++) {            printf("%c", u[i][j]);        }        printf("\n");    }    return 0;}