USACO－Runaround Numbers

最新推荐文章于 2018-04-18 20:55:26 发布

superxtong

最新推荐文章于 2018-04-18 20:55:26 发布

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分类专栏：杂题文章标签： UASCO

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本文介绍了一种特殊的整数——Runaround Numbers，这类数字的特点是所有位数都不相同且不为零，并通过特定的规则能够遍历每一位数字恰好一次。文章提供了一个C++实现示例，用于找出大于给定数字M的下一个Runaround Numbers。

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Runaround numbers are integers with unique digits, none of which is zero (e.g., 81362) that also have an interesting property, exemplified by this demonstration:

If you start at the left digit (8 in our number) and count that number of digits to the right (wrapping back to the first digit when no digits on the right are available), you’ll end up at a new digit (a number which does not end up at a new digit is not a Runaround Number). Consider: 8 1 3 6 2 which cycles through eight digits: 1 3 6 2 8 1 3 6 so the next digit is 6.
Repeat this cycle (this time for the six counts designed by the `6’) and you should end on a new digit: 2 8 1 3 6 2, namely 2.
Repeat again (two digits this time): 8 1
Continue again (one digit this time): 3
One more time: 6 2 8 and you have ended up back where you started, after touching each digit once. If you don’t end up back where you started after touching each digit once, your number is not a Runaround number.
Given a number M (that has anywhere from 1 through 9 digits), find and print the next runaround number higher than M, which will always fit into an unsigned long integer for the given test data.

PROGRAM NAME: runround

INPUT FORMAT

A single line with a single integer, M
SAMPLE INPUT (file runround.in)

81361
OUTPUT FORMAT

A single line containing the next runaround number higher than the input value, M.
SAMPLE OUTPUT (file runround.out)

81362

题意：给一串数字，要求每个数字不一样，每一次都走与数字对应的步数，问最终能否使每个数字都能被走一次且仅有一次。

模拟

代码：

/*
ID:iam666
PROG:runround
LANG:C++
*/

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <iostream>
using namespace std;
stack <int> save;
int work(unsigned long Num)
{
    int num[10];
    memset(num,0,sizeof(num));
    unsigned long tmp=Num;
    int le=0;
    while(tmp>0)
    {
        int i=tmp%10;
        save.push(i);
        num[i]++;
        if(num[i]>1)
            return 0;
        tmp/=10;
        le++;
    }
    int mark[le];
    for(int i=0;i<le;i++)
    {
        mark[i]=save.top();
        save.pop();
    }
    for(int i=0,ans=0;ans<le;ans++)
    {
        i+=mark[i];
        while(i>=le)
        {
            i-=le;
        }
        num[mark[i]]++;
    }
    for(int i=0;i<10;i++)
    {
        if(num[i]!=0&&num[i]!=2)
            return 0;
    }
    return 1;
}

int main (void)
{
    freopen("runround.in","r",stdin);
    freopen("runround.out","w",stdout);
    unsigned long n;
    cin>>n;
    unsigned long i=n+1;
    for(;work(i)==0;i++);
    printf("%lu\n",i);  
    return 0;


}