pat 1023. Have Fun with Numbers (20)

1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

解：存进位，存每个位数上的值，计算下就行了。

代码：

//Have Fun with Numbers
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    string s;
    cin>>s;
    int len=s.length();
    int *a=new int[len+1],*b=new int[len+1];
    int *temp1=new int[len+1],*temp2=new int[len+1];
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(temp1,0,sizeof(temp1));
    memset(temp2,0,sizeof(temp2));
    for(int i=0;i<len;i++)
    {
        a[i]=s[i]-48;
        temp1[i]=a[i];
    }
    sort(temp1,temp1+len);
    int flag=1,c=0;
    for(int i=len-1;i>=0;i--)
    {
        a[i]=a[i]*2+b[i];
        if(i>0)
            b[i-1]=a[i]/10;
        if(a[i]>9)
        {
            if(i==0)
                c=a[i]/10;
            a[i]=a[i]%10;
        }
    }
    for(int i=0;i<len;i++)
    {
        temp2[i]=a[i];
    }
    sort(temp2,temp2+len);
    for(int i=0;i<len;i++)
    {
        if(temp1[i]!=temp2[i])
        {
            flag=0;
            break;
        }
    }
    if(flag==1)
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
    if(c!=0) cout<<c;
    for(int i=0;i<len;i++)
    {
        cout<<a[i];
    }
    cout<<endl;
}