点击链接PAT甲级-AC全解汇总

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 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
1

Sample Output:

Yes
2469135798
1
2

题意：
输入一个数字，判断组成他的每一位数字能否组长他的两倍输出yes no。
无论可不可以，输出他的两倍。

我的代码：

#include
using namespace std;

int main()
{
    int record[10]={0};
    string input,output;
    cin>>input;
    for(auto it:input)record[it-'0']++;

    int jinwei=0;
    for(int i=input.length()-1;i>=0;i--)
    {
        int num=(input[i]-'0')*2+jinwei;
        jinwei=num/10;
        num%=10;
        record[num]--;
        output.insert(0,to_string(num));
    }
    if(jinwei)output.insert(0,"1");
    bool flag=true;
    for(auto it:record)if(it<0)flag=false;
    if(flag)cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
    cout<<output;
    return 0;
}

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