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\n");
103 } else
104 {
105 printf("Yes\n");
106 }
107 print(b);
108 return 0;
109 }