1023. Have Fun with Numbers (20)
时间限制
400 ms
内存限制
65536 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
原数由0-9几个数组成，看这个数乘二以后是否也由同样个数的0-9的排列组合。

 评测结果
时间结果得分题目语言用时(ms)内存(kB)用户7月30日 14:03答案正确201023C++ (g++ 4.7.2)1308datrilla
测试点
测试点结果用时(ms)内存(kB)得分/满分0答案正确13083/31答案正确11802/22答案正确13082/23答案正确13042/24答案正确13082/25答案正确13083/36答案正确13083/37答案正确13083/3
#include<iostream>     #include<string>using namespace std;  int main(){   int digits[10] = {0},index,c;  string strn,doublestr;  char tempc;  cin >> strn;   for (index = strn.size(),c=0; index > 0; )  {    index--;    digits[strn[index] - '0']++;    c += (strn[index] - '0')*2;    digits[c % 10]--;    tempc = c % 10 + '0';    doublestr = tempc + doublestr;    c = c / 10;  }  if (0 == c)  {    for (index = 0; index < 10&&0==c; index++)    {      if (0 != digits[index])c = 1;    }  }  else  {     tempc = c+ '0';    doublestr = tempc + doublestr;  }  if (0 == c)cout << "Yes" << endl;  else cout << "No" << endl;  cout << doublestr << endl;  system("pause");     return 0;}