A1023. Have Fun with Numbers (20)

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>#include<algorithm>using namespace std;int main(){bool flag = 1;//表示符合要求int hashTable[256] = {0};char str1[22],str2[22];scanf("%s",str1);int len1 = strlen(str1);int len2 = len1;//len2必须通过len1来求，否则会输出额外的东西？reverse(str1,str1+len1);//反转字符串for(int i = 0;i < len1;i++)hashTable[str1[i]]++;int carry = 0;for(int i = 0;i < len1;i++){str2[i] = ((str1[i]-'0')*2+carry)%10+'0';carry = ((str1[i]-'0')*2+carry)/10;}if(carry){ flag = 0;len2++;str2[len1] = carry+'0';}else {for(int i = 0;i < len1;i++)hashTable[str2[i]]--;for(int i = 0;i < len1;i++)if(hashTable[str2[i]]) flag = 0;}if(flag) printf("Yes\n");else printf("No\n");for(int i = len2-1;i >= 0;i--)printf("%c",str2[i]);printf("\n");return 0;}