1023. Have Fun with Numbers (20)
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> struct bign{ int d[21]; int len; bign(){ memset(d,0,sizeof(d)); len = 0; } }; bign change(char str[]){ bign a; a.len = strlen(str); for(int i = 0; i < a.len; i++){ a.d[i] = str[a.len- i - 1] - '0'; } return a; } bign multi(bign a, int b){ bign c; int carry = 0; for(int i = 0; i < a.len; i++){ int temp = a.d[i] * b + carry; c.d[c.len++] = temp%10; carry = temp/10; } while(carry != 0){ c.d[c.len++] = carry%10; carry /= 10; } return c; } bool judge(bign a,bign b){ if(a.len != b.len) return false; int count[10] = {0}; for(int i = 0; i < a.len; i++){ count[a.d[i]]++; count[b.d[i]]--; } for(int i = 0; i < 10; i++){ if(count[i] != 0) return false; } return true; } void print(bign a){ for(int i = a.len-1; i >= 0; i--){ printf("%d",a.d[i]); } } int main(){ char str[21]; gets(str); bign a = change(str); bign mul = multi(a,2); if(judge(a,mul) == true) printf("Yes\n"); else printf("No\n"); print(mul); return 0; }