1023. Have Fun with Numbers (20)
2016-01-23 20:09
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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:
Sample Output:
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
string doubleNum(string s){
string result;
int overflow = 0;
for(auto rit = s.rbegin(); rit != s.rend(); ++rit){
int num = (*rit - '0') * 2 + overflow;
result = char(num % 10 + '0') + result;
overflow = num / 10;
}
if(overflow) result = char(overflow + '0') + result;
return result;
}
int main(){
string s;
cin >> s;
string ss = doubleNum(s);
string out(ss);
sort(begin(s), end(s));
sort(begin(ss), end(ss));
cout << (s == ss ? "Yes" : "No") << endl;
cout << out;
return 0;
}
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
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
string doubleNum(string s){
string result;
int overflow = 0;
for(auto rit = s.rbegin(); rit != s.rend(); ++rit){
int num = (*rit - '0') * 2 + overflow;
result = char(num % 10 + '0') + result;
overflow = num / 10;
}
if(overflow) result = char(overflow + '0') + result;
return result;
}
int main(){
string s;
cin >> s;
string ss = doubleNum(s);
string out(ss);
sort(begin(s), end(s));
sort(begin(ss), end(ss));
cout << (s == ss ? "Yes" : "No") << endl;
cout << out;
return 0;
}
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