1023 Have Fun with Numbers

分数 20

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

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

#include<bits/stdc++.h>
using namespace std;map<int,int> se;
string add(string a,string b){vector<int> A,B,C;for(int i=a.size()-1;i>=0;i--) A.push_back(a[i]-'0');for(int i=b.size()-1;i>=0;i--) B.push_back(b[i]-'0');int t=0;for(int i=0;i<a.size()||i<b.size();i++){if(i<a.size()) t+=A[i];if(i<b.size()) t+=B[i];C.push_back(t%10);t/=10;}string c;if(t!=0) C.push_back(t);for(int i=C.size()-1;i>=0;i--){se[C[i]]--;if(se[C[i]]==0) se.erase(C[i]);c+=to_string(C[i]);}return c;
}int main(){string a;cin>>a;for(int i=0;i<a.size();i++) se[a[i]-'0']++;int digit = a.size();string b = add(a,a);if(se.size()==0) cout<<"Yes"<<endl;else cout<<"No"<<endl;cout<<b<<endl;}