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:

Yes
2469135798

#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;
}