[字符串处理]PAT1023 Have Fun with Numbers

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本文介绍了一种判断方法，用于检查一个给定的数字在乘以2后，其结果是否仅由原数字中的数字重新排列组成。通过示例输入输出展示程序的功能，并提供了一个C++实现方案。

1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

题意：求解一个长度为0~20之间的某个数乘2之后还是不是由原来的数字组成。

思路：直接模拟就可以，比较水。

#include<iostream>
#include<string>

using namespace std;

int arry[30],arry1[30],digit1[12],digit2[12];

bool same(int *st1,int *st2,int len)
{
    int i;
    for(i=0;i<len;i++)
    {
        digit1[st1[i]]++;
        digit2[st2[i]]++;
    }
    for(i=0;i<10;i++)
    {
        if(digit1[i]!=digit2[i]) return false;
    }
    return true;
}

int main()
{
    string str;
    cin>>str;
    int i,j,k=str.size();
    for(i=k-1,j=0;i>=0;i--,j++)
    {
        arry[j]=str[i]-'0';
    }
    for(i=0;i<k;i++)
    {
        arry1[i]=arry[i]*2;
    }
    int cnt=0;
    for(i=0;i<k;i++)
    {
        if(arry1[i]+cnt>=10)
        {
            int t=arry1[i]+cnt;
            arry1[i]=t%10;
            cnt=t/10;
        }
        else
        {
            arry1[i]=arry1[i]+cnt;
            cnt=0;
        }
    }
    if(cnt!=0) arry1[k++]=cnt;
    if(k!=str.size()) cout<<"No"<<endl;
    else
    {
        if(same(arry1,arry,k)) cout<<"Yes"<<endl;
        else cout<<"No"<<endl;
    }
    for(i=k-1;i>=0;i--) cout<<arry1[i];
    return 0;
}