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 <iostream>
2
3 #include <string>
4
5 #include <algorithm>
6
7 using namespace std;
8
9
10
11 int aa[21];
12
13 int v1[10];
14
15 int v2[10];
16
17
18
19 int main()
20
21 {
22
23
24
25 string ss;
26
27 while(cin>>ss)
28
29 {
30
31
32
33 int i;
34
35 for(i=0;i<21;i++)
36
37 {
38
39 aa[i]=0;
40
41 }
42
43 for(i=0;i<10;i++)
44
45 {
46
47 v1[i]=0;
48
49 v1[i]=0;
50
51 }
52
53 int count=0;
54
55 for(i=ss.length()-1;i>=0;i--)
56
57 {
58
59 aa[count++]=ss[i]-‘0‘;
60
61 v1[ss[i]-‘0‘]=1;
62
63 }
64
65
66
67 for(i=0;i<count;i++)
68
69 {
70
71 aa[i]=aa[i]*2;
72
73 }
74
75
76
77 int tem,len;
78
79 for(i=0;i<count;i++)
80
81 {
82
83 if(aa[i]>9)
84
85 {
86
87 tem=aa[i]/10;
88
89 aa[i+1]=aa[i+1]+tem;
90
91 aa[i]=aa[i]%10;
92
93 }
94
95 }
96
97
98
99 if(aa[count]==0) len=count;
100
101 else len=count+1;
102
103
104
105
106
107 reverse(aa,aa+len);
108
109
110
111 for(i=0;i<len;i++)
112
113 {
114
115 v2[aa[i]]=1;
116
117 }
118
119 bool ifis=true;
120
121 for(i=0;i<10;i++)
122
123 {
124
125 if(v1[i]!=v2[i]) ifis=false;
126
127 }
128
129
130
131 if(ifis) cout<<"Yes"<<endl;
132
133 else cout<<"No"<<endl;
134
135
136
137 for(i=0;i<len;i++)
138
139 {
140
141 cout<<aa[i];
142
143 }
144
145 cout<<endl;
146
147
148
149 }
150
151 return 0;
152
153 }