• 九度oj 1464 Hello World for U 2012年浙江大学计算机及软件工程研究生机试真题


    题目1464:Hello World for U

    时间限制:1 秒

    内存限制:128 兆

    特殊判题:

    提交:3872

    解决:1082

    题目描述:

    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.

    输入:

    There are multiple test cases.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.

    输出:

    For each test case, print the input string in the shape of U as specified in the description.

    样例输入:
    helloworld!
    ac.jobdu.com
    样例输出:
    h   !
    e   d
    l   l
    lowor
    a    m
    c    o
    .    c
    jobdu.
    来源:
    2012年浙江大学计算机及软件工程研究生机试真题
    分析:
     1 #include<iostream>
     2 #include<queue>
     3 #include<cstdio>
     4 #include<cstring>
     5 #include<cmath>
     6 #include<algorithm>
     7 using namespace std;
     8 int main(){
     9     string s;
    10     int n1,n3;
    11     while(cin>>s){
    12         int len=s.length()-1;
    13         int n=len+1;
    14         int nn=n+2;
    15         while(nn%3){
    16             nn--;
    17         }
    18         n1=nn/3;
    19         //cout<<n1<<endl;
    20         n3=n-n1*2+2;
    21         int i=0;
    22         //cout<<n3<<endl;
    23         for(;i<n1-1;i++){
    24             cout<<s[i];
    25             int j=0;
    26             for(;j<n3-2;j++){
    27                 cout<<" ";
    28                 
    29             }
    30             cout<<s[len-i]<<endl;
    31         }
    32         int k=i+n3;
    33         for(;i<k;i++){
    34             cout<<s[i];
    35         }
    36         cout<<endl;
    37     }
    38     return 0;
    39 }
  • 相关阅读:
    深入研究Node.js的底层原理和高级使用
    Node.js实际案例分享
    Node.js的下载、安装、配置、Hello World、文档阅读
    Node.js初识
    sphinx初识
    Yii2中后台用前台的代码设置验证码显示不出来?
    彻底解决Yii2中网页刷新时验证码不刷新的问题
    yii2高级模板使用一个域名管理前后台
    Shot(三分)
    遮挡判断(模拟)
  • 原文地址:https://www.cnblogs.com/Deribs4/p/4288979.html
Copyright © 2020-2023  润新知