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

#include <cstdio>#include <cstring>using namespace std;void doubleNum(char *s){   int c = 0;   int len = strlen(s);   int sum[100];   memset(sum, 0, sizeof(sum));   int cnt = 0;   int num[10];    memset(num, 0, sizeof(num));   for(int i = 0; i < len; i++)   {      num[s[i] - '0']++;   }   int flag = 1;   for(int i = len - 1; i >= 0; i--)   {        sum[cnt++] = ((s[i] -'0')* 2 + c) % 10;        c =  ((s[i] -'0')* 2 + c) /10;        num[sum[cnt - 1]]--;   }   if(c > 0)    sum[cnt++] = c;   for(int i = 0; i < 10; i++)        if(num[i] != 0)        {           flag = 0;        }    if(flag == 1)        printf("Yes\n");    else{        printf("No\n");    }   for(int i = cnt - 1; i >= 0; i--)   {      printf("%d",sum[i]);   }}int main(){     char s[100];     scanf("%s", s);     doubleNum(s);}