1023 Have Fun with Numbers

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：

　　将一个数字翻倍后，判断该数字是不是原来数字的重新排列。

思路：

　　大数加法。

Code：

#include <bits/stdc++.h>

using namespace std;

int main() {
    string s;
    cin >> s;
    int len = s.length();
    int carry = 0;
    map<int, int> m;
    for (int i = 0; i < s.length(); ++i) m[s[i] - '0']++;
    for (int i = len - 1; i >= 0; --i) {
        s[i] += s[i] + carry - '0';
        carry = 0;
        if (s[i] > '9') {
            s[i] -= 10;
            carry = 1;
        }
    }
    if (carry == 1) s = '1' + s;
    int flag = 0;
    for (int i = 0; i < s.length(); ++i)
        if (m[s[i] - '0'] > 0)
            m[s[i] - '0']--;
        else
            flag = 1;
    if (flag == 0)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    cout << s << endl;
    return 0;
}