1023 Have Fun with Numbers (20)（20 分）

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1 题目

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

题目链接：

2 解题思路

　　给定一个数，乘以2之后，各个数位的出现次数，是否与乘之前是否相同，相同输出Yes，否认输出No，然后下一行打印出乘2之后的各个数位。
　　这个题目是属于超大数运算的类型，题目给出的最高数位是20位，那么即便是用最高的unsigned long long 也会面临溢出的情况，所以输入和输出，只能用string，诸位乘2，然后再记录每一位出现的次数，相比较就行。

3 AC代码

/*
** @Brief:No.1023 of PAT advanced level.
** @Author:Jason.Lee
** @Date:2018-8-04
*/

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;

int digit[10];
string input;
string output;

bool doubleNumber(string input){
    int flag = 0;
    int temp = 0;
    output.resize(input.length());
    for(int i = input.length()-1;i>=0;i--){
        temp = (int)(input[i]-'0');

        digit[temp]++;
        temp=2*temp+flag;

        if(temp>9){
            temp%=10;
            flag = 1;
        }else{
            flag = 0;
        }
        output[i] = (temp+'0');
        //cout<<"ouput ："<<output[i]<<endl;
    }
    if(flag){
        output= "1"+output;
    }
    //cout<<"output : "<<output<<endl;
    //cout<<"size : "<<output.length()<<endl;
    for(int i = 0;i<output.length();i++){
        digit[(int)(output[i]-'0')]--;
    }
    for(int i=0;i<10;i++){
        if(digit[i]!=0){
            return false;
        }
    }
    return true;
}
int main(){
    cin>>input;
    cout<<(doubleNumber(input)? "Yes":"No")<<endl;
    cout<<output<<endl;
    return 0;
}