HDU 2991 || SDUT 2394 Generate random numbers(简单数论)

Generate random numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 114    Accepted Submission(s): 44

Problem Description

John von Neumann suggested in 1946 a method to create a sequence of pseudo-random numbers. His idea is known as the "middle-square"-method and works as follows: We choose an initial value a₀, which has a decimal representation of length at most n. We then multiply the value a₀ by itself, add leading zeros until we get a decimal representation of length 2 × n and take the middle n digits to form a_i. This process is repeated for each a_i with i>0. In this problem we use n = 4. 

Example 1: a₀=5555, a₀²=30858025, a₁=8580,... 

Example 2: a₀=1111, a₀²=01234321, a₁=2343,... 

Unfortunately, this random number generator is not very good. When started with an initial value it does not produce all other numbers with the same number of digits. 

Your task is to check for a given initial value a₀ how many different numbers are produced. 

 

Input

The input contains several test cases. Each test case consists of one line containing a₀ (0 < a₀ < 10000). Numbers are possibly padded with leading zeros such that each number consists of exactly 4 digits. The input is terminated with a line containing the value 0. 

Output

For each test case, print a line containing the number of different values a_i produced by this random number generator when started with the given value a₀. Note that a₀ should also be counted.

 

Sample Input

5555

0815

6239

0

 

Sample Output

32

17

111

 

Hint

 

Note that the third test case has the maximum number of different values among all possible inputs.

Source

2009/2010 Ulm Local Contest

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解题报告：这道题主要是题意不太好懂，今天内部测试的时候花了很长时间才看懂，哎！英语不行呀，这道题就是给我们四位数（形式上是四位，可以有前导零例如：0021），让它产生随机数，方法是让其平方后为形式上的八位数例如：0021平方后为00000441去中间的四位0004，依此类推，终究会有一个四位数和前面的一样，就是求这个数的位置；

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1010;
int a[N], a2[N];//a[]存储的是“形式上”的四位数，a2[]存储的是“形式上”的八位数
int main()
{
    int n, i, ans , j, flag;
    while (scanf("%d", &n) != EOF && n)
    {
        memset(a, 0, sizeof(a));
        memset(a2, 0, sizeof(a2));
        a[0] = n;
        for (i = 0; i < N; ++i)
        {
            a2[i] = a[i] * a[i];
            a[i + 1] = a2[i] / 100 - (a2[i] / 1000000) * 10000; //取中间的四位数
        }
        flag = 0;
        for (i = 0; i < N && !flag; ++i)
        {
            for (j = i + 1; j < N && !flag; ++j)
            {
                if (a[i] == a[j])//找到第一个四位数和之前的一样
                {
                    flag = 1;
                    ans = j;//记录这个数的位置
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}