[ PAT-A ] 1023 Have Fun with Numbers (C++)

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这是一个关于计算机算法的问题，要求判断一个不超过20位的正整数，当该数翻倍后，其数字是否仍为原数的排列。输入一个正整数，输出'Yes'或'No'并显示翻倍结果。

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题目描述

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

解题思路

题目大意
给出一个长度不超过20的整数，问这个整数两倍后的数位是否为原数位的一个排列。不管是yes还是no最后都要输出整数乘以2的结果
思路
将这一串数字以字符形式存储至数组中，进行倍数操作时，注意最高位即可

代码设计

[任务] 将一串数字乘以２，并判断是否可由原先数字重新排列得出
[接口]
bool doubler(char ss[],char* re,int& ans)
输入:ss 一串数字　re 存放乘以２后的结果(如果数字长度超过ss的长度，将最高位放置ans中)　ans 存放re的最高位

//代码实现
//zhicheng
bool doubler(char ss[],char* re,int& ans)
{
    int book1[10]={0},book2[10]={0};
    int len=strlen(ss);
    re[len]='\0';
    for(int i=len-1;i>=0;i--)
    {
        ans+=(ss[i]-'0')*2;
        re[i]=ans%10+'0';
        ans/=10;
        book1[ss[i]-'0']++;
        book2[re[i]-'0']++;
    }
    if(ans)book2[ans]++;
    for(int i=0;i<10;i++)if(book1[i]!=book2[i])return false;
    return true;
}