7-8 Have Fun with Numbers （20 分）
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 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
//数字出现的次数输入前与变换后是否相等，检查范围是【0,9】；
//数字的总位数输入前与输入后是否相等；
//易错样例：59，第一位数2之后大于等于10；应输出No 118;
//输入53输出No 106;
//输入999输出No 1998；
#include<stdio.h>
#include<string.h>
int main()
{
char ch[1000];
int i,j,a[1000]={0},temp,k=0,n1[1000]={0},n2[1000]={0},flag=0,temp1,len=0;
gets(ch);
len=strlen(ch);//原来数字的位数；
for(i=0;ch[i]!=’\0’;i++)
{
temp=ch[i]-‘0’;
temp1=temp2;
n1[temp]++;//统计原始数字出现的次数；
if(temp1<10) a[k++]=temp1%10;
else if(temp1>=10)
{
if(i= =0)//输入59或53；
{
a[k]=temp1/10;//把十位数赋值给a[0];
k++;
a[k]=temp1%10;//把个位数赋值给a[1];
k++;
}
else if(i>=1)
{
a[k-1]=a[k-1]+temp1/10;//向高位进位；
a[k++]=temp1%10;//保留了进位之后的数字
}
}
}
for(i=0;i<k;i++)//k是变换之后数字的位数；
{
n2[a[i]]++;//变换之后的数字出现的次数；
}
for(i=0;i<=9;i++)//若输入以5结尾的数1234567895；则二倍之后末尾是零；
{
if(n1[i]!=n2[i])//比较一下原来的数字和变换之后的数字出现得次数是否相同；
{
flag=1;
break;
}
}
if(flag= =0&&k==len)
{
printf(“Yes\n”);
for(i=0;i<k;i++)
{
printf("%d",a[i]);
}
}
else
{
printf(“No\n”);
for(i=0;i<k;i++)
{
printf("%d",a[i]);
}
}
return 0;
}