1023 Have Fun with Numbers简单的大整数乘法

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本文探讨了一个有趣的数学问题，即检查一个给定的数字，在乘以2后，其结果是否仅由原数字的位数排列组成。通过算法实现，验证了这一特性，并提供了详细的代码解释。

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1023 Have Fun with Numbers （20 分）
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 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

题目大意：给你一个20个数字以内的整数，让你判断这个数乘以2之后与原来的数相比，1-9这9个数字出现的次数是不是一样的，是则输出Yes，否则No，最后输出这个数乘以2的结果

#include <cstdio>
#include <algorithm>

using namespace std;

int main()
{
    int a[21], b[21], dou[21], len=0;
    char c;
    while(scanf("%c",&c))
    {
        if (c == '\n')
            break;
        a[len++] = int(c-'0');
    }
    int tmp = 0;
    for (int i=len-1; i>=0; i--)
    {
        b[i] = 2*a[i]+tmp;
        tmp = b[i] / 10;
        b[i] %= 10;
        dou[i] = b[i];
    }
    if (tmp != 0)
        printf("No\n1");
    else
    {
        bool equ = true;
        sort(a,a+len);
        sort(b,b+len);
        for (int i=0; i<len; i++)
        {
            if (a[i] != b[i])
                equ = false;
        }
        if (equ == true)
            printf("Yes\n");
        else
            printf("No\n");
    }
    for (int i=0; i<len; i++)
        printf("%d",dou[i]);
    return 0;
}