Have Fun with Numbers (大数)

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 }

View Code

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原文地址：https://www.cnblogs.com/xiaoyesoso/p/4240611.html