7-49 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.
用例输入:
1234567899
输出
Yes
2469135798
细节注意点:1.输出No的时候也要输出数字;2.最高位进位要显示
#include<stdio.h>
#include<string.h>
int main()
{
char num[21];
gets(num);
int a[10],b[10];
for(int i=0;i<10;i++)
{
a[i]=b[i]=0;
}
int flag=1;
for(int i=0;i<strlen(num);i++)
{
a[num[i]-'0']++;
}
char tmp[21];
int jinwei=0;
for(int i=strlen(num)-1;i>=0;i--)
{
tmp[i]=((num[i]-'0')*2+jinwei)%10+'0';
if((num[i]-'0')*2>9)
{
jinwei=1;
}
else
{
jinwei=0;
}
}
tmp[strlen(num)]=0;
for(int i=0;i<strlen(num);i++)
{
b[tmp[i]-'0']++;
}
for(int i=0;i<10;i++)
{
if(a[i]!=b[i])
{
flag=0;
break;
}
}
if(flag==1)
{
printf("Yes\n");
printf("%s",tmp);
}
else
{
printf("No\n");
if(jinwei==1)
{
printf("1%s",tmp);
}
else
{
printf("%s",tmp);
}
}
return 0;
}