自测4. 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

分析：这道题的意思呢，就是给一个不超过20位的数，然后乘以2，如果之后得到的数的组成与之前一样，顺序不同，那么就输出“Yes”，否则就“No”咯。要声明一下，这道题如果用长整形之类的话，那么大数的乘法是会有误差的，可以自己先试试，你会发现得到的数会让你大吃一惊。所以我们需要以另外一种方式来解决，我将每一位作为一个字符输入，然后再转化为整形数，存到一个整形数组中，接着通过两个存储了乘2前后数组成的数组比较，来判断是否相同，由此来输出答案，当然这种方法需要你自己写代码来进行乘法运算。只是乘2，应该都会得。

code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
char c;
int cnt=0;    //计数器
//输入数字
int num[20];
int a[10]={0};
int b[10]={0};  //用于比较的数组a,b，其中元素为含有这个数的个数
while(1)
{
scanf("%c",&c);
if((int)c-48>=0 && (int)c-48<=9)
{
num[cnt] = (int)c-48;       //将数存到数组中
a[num[cnt]]++;   //个数+1
cnt++;
}
else
break;
}
//对数进行*2处理
int i;
int flag =0; //进位
for(i=cnt-1;i>=0;i--)   //从最低位开始
{
num[i] = num[i]*2+flag;     //*2后的数
if(num[i]>9 && i!=0)
{
flag =1;    //进位为1
num[i] = num[i]%10; //只留个位
}
else
flag =0;
if(num[0]<=9)
b[num[i]]++;      //个数+1
}
//判断a，b数组是否相同
int sign = 0;
for(i=0;i<10;i++)
{
if(a[i]!=b[i])
{
sign = 1;
break;
}
}
//输出
if(sign == 0)
printf("Yes\n");
else
printf("No\n");
for(i=0;i<cnt;i++)
{
printf("%d",num[i]);
}
return 0;
}