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
 

題目解析
　　本題給出一個長度再20以內的數字（由1~9組成），要求判斷這個數字加倍後的新數字是不是這個數字的某一種排列，如果是的化輸出Yes否則輸出No，之後輸出加倍後的數字。

　　由於數字最大位數為20位，超過了long long int的記錄範圍我們用數組num記錄這個數字，用數組cnt記錄num中1~9出現的次數，將num加倍後判斷其中1~9出現的次數是否發送改變，若沒有發送改變則證明加倍後的數字是原數字的某一種排列，反之則不是。

AC代碼

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int num[22];
 4 int cnt[10];
 5 string str;
 6 void toInt(){
 7     for(int i = 0; i < str.size(); i++)
 8         num[i] = str[i] - ‘0‘;
 9 }
10 void getCnt(){
11     for(int i = 0; i < str.size(); i++)
12         cnt[num[i]]++;
13 }
14 bool judge(int carry){
15     if(carry != 0)  //如果最高位進位不為零，則證明加倍後的數字比原數字多一位，那麽其肯定不是原數字的一個排列
16         return false;
17     for(int i = 0; i < str.size(); i++)
18         cnt[num[i]]--;
19     for(int i = 1; i <= 9; i++){    //判斷新的num中1~9的數量是否和加倍前一樣
20         if(cnt[i] != 0)
21             return false;
22     }
23     return true;
24 }
25 int doubleNumber(){ //將數組num加倍並返回最高位進位
26     int carry = 0;
27     for(int i = str.size() - 1; i >= 0; i--){
28         int temp = num[i];
29         num[i] = (2 * temp + carry) % 10;
30         carry = 2 * temp / 10;
31     }
32     return carry;
33 }
34 int main()
35 {
36     cin >> str; //輸入數字
37     toInt();    //將輸入的數字轉化為數組
38     getCnt();   //獲取數組中1~9出現的次數
39     int carry = doubleNumber(); //將num加倍carry記錄最高位的進位
40     if(judge(carry)){   //判斷加倍後的數字是否為原數字的某一個排列
41         printf("Yes\n");
42         
43     }else
44         printf("No\n");
45     if(carry != 0)  //判斷是否需要輸出進位
46         printf("%d", carry);
47     for(int i = 0; i < str.size(); i++) //輸出加倍後的數組num
48         printf("%d", num[i]);
49     printf("\n");
50     return 0;
51 }

PTA (Advanced Level)1023 Have Fun with Numbers