• 1031 Hello World for U (20 分)(图形打印)【回顾】


    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   !
    e   d
    l   l
    lowor
    

    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
    

    生词

    英文 解释
    vertical 垂直的
    squared 方形的

    题目大意:

    用所给字符串按U型输出。n1和n3是左右两条竖线从上到下的字符个数,n2是底部横线从左到右的字符个数。
    要求:

    1. n1 == n3
    2. n2 >= n1
    3. n1为在满足上述条件的情况下的最大值

    分析:

    假设n = 字符串长度 + 2,因为2 * n1 + n2 = n,且要保证n2 >= n1, n1尽可能地大,分类讨论:

    1. 如果n % 3 == 0,n正好被3整除,直接n1 == n2 == n3;

    2. 如果n % 3 == 1,因为n2要比n1大,所以把多出来的那1个给n2

    3. 如果n % 3 == 2, 就把多出来的那2个给n2

    所以得到公式:n1 = n / 3,n2 = n / 3 + n % 3

    把它们存储到二维字符数组中,一开始初始化字符数组为空格,然后按照u型填充进去,最后输出这个数组u。

    原文链接:https://blog.csdn.net/liuchuo/article/details/52119867

    题解

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

    本文来自博客园,作者:勇往直前的力量,转载请注明原文链接:https://www.cnblogs.com/moonlight1999/p/15536335.html

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  • 原文地址:https://www.cnblogs.com/moonlight1999/p/15536335.html
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