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
Sample Output:
Yes
2469135798
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<bits/stdc++.h>
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;
}