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 kdigits, 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
直接用字符串读入更好处理。
1 #include <bits/stdc++.h> 2 using namespace std; 3 int vis[10]; 4 string s; 5 bool flag = true; 6 7 int main(){ 8 cin >> s; 9 for(int i = 0; i < s.length(); i++){ 10 int x = s[i] - '0'; 11 vis[x]++; 12 } 13 int ans = 0; 14 for(int i = s.length()-1; i >= 0; i--){ 15 int x = s[i] - '0'; 16 int k = x*2 + ans; 17 int an = k%10; 18 s[i] = an + '0'; 19 if(vis[an] >= 1) 20 vis[an]--; 21 else{ 22 flag = false; 23 } 24 ans = k/10; 25 } 26 if(ans == 1) 27 s = '1'+ s; 28 if(flag) 29 cout <<"Yes"<<endl; 30 else 31 cout <<"No"<<endl; 32 cout <<s<<endl; 33 return 0; 34 }