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: Yes 2469135798
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){ int a[25],b[25]; char str[25],str2[25]; scanf("%s",str); int len=strlen(str); for(int i=0;i<len;i++){ a[i]=str[len-1-i]-'0'; } int i,carry=0; for(i=0;i<len;i++){ b[i]=(a[i]*2+carry)%10; carry=(a[i]*2+carry)/10; } if(carry){ b[i++]=carry; } int b_len=i; for(int i=0;i<b_len;i++){ str2[i]=b[b_len-1-i]+'0'; } str2[b_len]='/0';//切記不能忘記字符串結(jié)束字符 sort(str,str+len); sort(str2,str2+b_len);// printf("%s %s/n",str,str2); if(strcmp(str,str2)==0){ printf("Yes/n"); }else{ printf("No/n"); } for(int i=b_len-1;i>=0;i--){ printf("%d",b[i]); } printf("/n"); return 0; }新聞熱點(diǎn)
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