PAT1023. Have Fun with Numbers

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

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使用高级编辑器

代码

1

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 int data[10]={
 6     0
 7 };
 8 struct Bignum
 9 {
10     int d[1000];
11     int len;
12     Bignum()  //构造函数
13     {
14         memset(d,0,sizeof(d));
15         len=0;
16     }
17 };
18 Bignum A;
19 void Mul()
20 {
21     int carry=0;
22     for(int i=0;i<A.len;i++)
23     {
24         A.d[i]=A.d[i]*2+carry;
25         if(A.d[i]>=10)
26         {
27             carry=A.d[i]/10;
28             A.d[i]%=10;
29         }
30         else
31             carry=0;
32
33     }
34     if(carry!=0)  //此处有错误
35     {
36         A.len++;
37         A.d[A.len-1]=carry%10;
38         carry/=10;
39     }
40 }
41 int main(int argc, char *argv[])
42 {
43
44     char ori[100];
45     scanf("%s",ori);
46     int length=strlen(ori);
47     for(int i=length-1;i>=0;i--)
48     {
49         A.d[A.len++]=ori[i]-‘0‘;
50         data[ori[i]-‘0‘]++;
51     }
52     Mul();
53     bool flag=true;
54     for(int i=A.len-1;i>=0;i--)
55     {
56         data[A.d[i]]--;
57     }
58     for(int i=0;i<10;i++)
59     {
60         if(data[i]!=0)
61         {
62             flag=false;
63             break;
64         }
65     }
66     if(flag)
67       printf("Yes\n");
68     else
69       printf("No\n");
70     for(int i=A.len-1;i>=0;i--)
71     {
72         printf("%d",A.d[i]);
73     }
74     putchar(‘\n‘);
75     return 0;
76 }