【PAT (Advanced Level)】1023. Have Fun with Numbers (20)

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1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

1234567899

Sample Output:

Yes2469135798

输入一个20为以内的数，把它加倍，如果得到的新数中所包含0~9的个数和原来一样，我们就输入“Yes"，否则输出”No"；两个数组，一个用来存放新数，另一个用来记录0~9的个数；

代码如下：

#include <iostream>#include <string>using namespace std;int main(){  string s;  while (cin >> s)  {    int cNum[10] = {0};    int dNum[21] = {0};    int t = 0;    for (int i = s.length(); i > 0; i--)    {      cNum[s[i - 1] - '0']++;      dNum[i - 1] = ((s[i - 1] - '0') * 2 + t) % 10;      t = ((s[i - 1] - '0') * 2 + t) / 10;    }    for (int i = s.length(); i > 0; i--)    {      cNum[dNum[i - 1]]--;    }    for (int i = 0; i < 10; i++)    {      if (cNum[i] > 0)      {        cout << "No" << endl;    break;      }      if (i == 9) cout << "Yes" << endl;    }    if (t != 0) cout << t;    for (int i = 0; i < s.length(); i++)    {      cout << dNum[i];    }    cout << endl;  }  return 0;}

因为这次是在网页上直接写的，所以就没有了Enjoy it！的注释了= =

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