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
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <iostream> 4 #include <string.h> 5 #include <string> 6 #include <math.h> 7 #include <algorithm> 8 using namespace std; 9 10 struct bign 11 { 12 int d[1010]; 13 int len; 14 bign(){ 15 memset(d,0,sizeof(d)); 16 len=0; 17 } 18 }; 19 20 bign change(char str[]){ 21 bign a; 22 a.len=strlen(str); 23 for(int i=0;i<a.len;i++) 24 { 25 a.d[i]=str[a.len-i-1]-'0'; 26 } 27 return a; 28 } 29 30 bign divide(bign a,int b,int & r) 31 { 32 bign c; 33 c.len=a.len; 34 for(int i=a.len-1;i>=0;i--) 35 { 36 r=10*r+a.d[i]; 37 if(r<b)c.d[i]=0; 38 else 39 { 40 c.d[i]=r/b; 41 r=r%b; 42 } 43 } 44 while(c.len-1>0&&c.d[c.len-1]==0) 45 { 46 c.len--; 47 } 48 return c; 49 } 50 51 52 bign multi(bign a,int b) 53 { 54 bign c; 55 int carry=0; 56 for(int i=0;i<a.len;i++) 57 { 58 int temp; 59 temp=a.d[i]*b+carry; 60 carry=temp/10; 61 c.d[c.len++]=temp%10; 62 } 63 //乘法高位处理 64 while(carry!=0) 65 { 66 c.d[c.len++]=carry%10; 67 carry/=10; 68 } 69 return c; 70 } 71 72 void print(bign a) 73 { 74 for(int i=a.len-1;i>=0;i--) 75 { 76 printf("%d",a.d[i]); 77 } 78 } 79 80 bool judge(bign a ,bign b) 81 { 82 if(a.len!=b.len)return false; 83 int count[10]={0}; 84 for(int i=0;i<a.len;i++ ) 85 { 86 count[a.d[i]]++; 87 count[b.d[i]]--; 88 } 89 for(int i=0;i<10;i++) 90 { 91 if(count[i]!=0)return false; 92 } 93 return true; 94 } 95 int main(){ 96 char str[21]; 97 scanf("%s",str); 98 bign a=change(str); 99 bign b=multi(a,2); 100 if(judge(a,b)==false) 101 { 102 printf("No "); 103 } else 104 { 105 printf("Yes "); 106 } 107 print(b); 108 return 0; 109 }