poj 2183 Bovine Math Geniuses_不见不散的结局是曲终人散_新浪博客

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2532		Accepted: 1574

Description

Farmer John loves to help the cows further their mathematical skills. He has promised them Hay-flavored ice cream if they can solve various mathematical problems. 

He said to Bessie, "Choose a six digit integer, and tell me what it is. Then extract the middle four digits. Square them and discard digits at the top until you have another number six digits or shorter. Tell me the result." 

Bessie, a mathematical genius in disguise, chose the six digit number 655554. "Moo: 6 5 5 5 5 4", she said. She then extracted the middle four digits: 5555 and squared them: 30858025. She kept only the bottom six digits: 858025. "Moo: 8 5 8 0 2 5", she replied to FJ. 

FJ nodded wisely, acknowledging Bessie's prowess in arithmetic. "Now keep doing that until you encounter a number that repeats a number already seen," he requested. 

Bessie decided she'd better create a table to keep everything straight: 

              Middle    Middle   Shrunk to


       Num   4 digits   square   6 or fewer


     655554    5555    30858025    858025


     858025    5802    33663204    663204


     663204    6320    39942400    942400


     942400    4240    17977600    977600


     977600    7760    60217600    217600 <-+


     217600    1760     3097600     97600   |


      97600    9760    95257600    257600   |


     257600    5760    33177600    177600   |


     177600    7760    60217600    217600 --+

Bessie showed her table to FJ who smiled and produced a big dish of delicious hay ice cream. "That's right, Bessie," he praised. "The chain repeats in a loop of four numbers, of which the first encountered was 217600. The loop was detected after nine iterations." 

Help the other cows win ice cream treats. Given a six digit number, calculate the total number of iterations to detect a loop, the first looping number encountered, and also the length of the loop. 

FJ wondered if Bessie knew all the tricks. He had made a table to help her, but she never asked: 

              Middle    Middle   Shrunk to


       Num   4 digits   square   6 or fewer


     200023    0002       4        4


          4       0       0        0


          0       0       0        0         [a self-loop]

whose results would be: three iterations to detect a loop, looping on 0, and a length of loop equal to 1. 

Remember: Your program can use no more than 16MB of memory. 

Input

* Line 1: A single six digit integer that is the start of the sequence testing. 

Output

* Line 1: Three space-separated integers: the first number of a loop, the length of the loop, and the minimum number of iterations to detect the loop. 

Sample Input

655554

Sample Output

217600 4 9

题意：输入一个六位数N，去掉头尾，取中间四位组成一个新的四位数N1，然后将N1平方后取后六位数N2，此上记录为操作K,然后从N2又重复以上的操作K，如此循环。求第一次遇到数六位数M到第二次遇到数M需要的操作K的次数。并求出总得操作K的次数；

思路：需要将当前所得到的六位数存放在数组，遇到重复的数字，输出即可；

代码如下：

#include < cstdio > 

#include < cstring >

#include < ctime >

int s[1000000];

int ans=0,cnt=0;

void dfs(int m)

{

    s[cnt++]=m;

    m=m % 100000;//去掉首位数；

    m=(m-m % 10000 % 1000 % 100 % 10)/10;//去掉尾位数；

    m=m*m;

    m=m % 10000000 % 1000000;//取后六位数；

    ans++;

    for(int i=0;i < cnt;i++)

    {

        if(m==s[i])

        {

           printf("%d %d %d\n",m,ans-i,ans);

           return ;

        }

    }

    dfs(m);

}

int main()

{

    int m;

    scanf("%d",&m);

        dfs(m);

    return 0;

}