zoukankan      html  css  js  c++  java
  • HDU6154

    CaoHaha's staff

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 28    Accepted Submission(s): 17


    Problem Description

    "You shall not pass!"
    After shouted out that,the Force Staff appered in CaoHaha's hand.
    As we all know,the Force Staff is a staff with infinity power.If you can use it skillful,it may help you to do whatever you want.
    But now,his new owner,CaoHaha,is a sorcerers apprentice.He can only use that staff to send things to other place.
    Today,Dreamwyy come to CaoHaha.Requesting him send a toy to his new girl friend.It was so far that Dreamwyy can only resort to CaoHaha.
    The first step to send something is draw a Magic array on a Magic place.The magic place looks like a coordinate system,and each time you can draw a segments either on cell sides or on cell diagonals.In additional,you need 1 minutes to draw a segments.
    If you want to send something ,you need to draw a Magic array which is not smaller than the that.You can make it any deformation,so what really matters is the size of the object.
    CaoHaha want to help dreamwyy but his time is valuable(to learn to be just like you),so he want to draw least segments.However,because of his bad math,he needs your help.
     

    Input

    The first line contains one integer T(T<=300).The number of toys.
    Then T lines each contains one intetger S.The size of the toy(N<=1e9).
     

    Output

    Out put T integer in each line ,the least time CaoHaha can send the toy.
     

    Sample Input

    5 1 2 3 4 5
     

     

    Sample Output

    4 4 6 6 7
     

    Source

     
    贪心
     1 //2017-08-19
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <iostream>
     5 #include <algorithm>
     6 #include <cmath>
     7 
     8 using namespace std;
     9 
    10 int main()
    11 {
    12     int T, n;
    13     scanf("%d", &T);
    14     while(T--){
    15         scanf("%d", &n);
    16         int len = sqrt(n/2);
    17         int area = 2*len*len;
    18         if(n-area == 0){
    19             printf("%d
    ", len*4);
    20         }else if(n <= area+len-1){
    21             printf("%d
    ", len*4+1);
    22         }
    23         else if(n-area <= 2*len){
    24             printf("%d
    ", len*4+2);
    25         }else if(n <= 2*(len+1)*(len+1)-(len+1+1)){
    26             printf("%d
    ", len*4+3);
    27         }
    28         else{
    29             printf("%d
    ", (len+1)*4);
    30         }
    31     }
    32 
    33     return 0;
    34 }
  • 相关阅读:
    第27章 结构型模式大PK
    第26章 创建型模式大PK
    Linux下C/C++代码调用PHP代码(转)
    php 5.2.17 升级到5.3.29
    PHP安全之register_globals (转)
    注意字符串的strlen与sizeof的差别
    嵌入式串口传送汉字乱码
    字符设备驱动模块与测试代码编写(转)
    字符设备之register_chrdev与register_chrdev_region(转)
    I2C子系统驱动框架及应用 (转)
  • 原文地址:https://www.cnblogs.com/Penn000/p/7397324.html
Copyright © 2011-2022 走看看