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:1234567899Sample Output:
Yes 2469135798
题意：求解一个长度为0~20之间的某个数乘2之后还是不是由原来的数字组成。
思路：直接模拟就可以，比较水。
#include<iostream>
#include<string>
using namespace std;
int arry[30],arry1[30],digit1[12],digit2[12];
bool same(int *st1,int *st2,int len)
{
int i;
for(i=0;i<len;i++)
{
digit1[st1[i]]++;
digit2[st2[i]]++;
}
for(i=0;i<10;i++)
{
if(digit1[i]!=digit2[i]) return false;
}
return true;
}
int main()
{
string str;
cin>>str;
int i,j,k=str.size();
for(i=k-1,j=0;i>=0;i--,j++)
{
arry[j]=str[i]-‘0‘;
}
for(i=0;i<k;i++)
{
arry1[i]=arry[i]*2;
}
int cnt=0;
for(i=0;i<k;i++)
{
if(arry1[i]+cnt>=10)
{
int t=arry1[i]+cnt;
arry1[i]=t%10;
cnt=t/10;
}
else
{
arry1[i]=arry1[i]+cnt;
cnt=0;
}
}
if(cnt!=0) arry1[k++]=cnt;
if(k!=str.size()) cout<<"No"<<endl;
else
{
if(same(arry1,arry,k)) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
for(i=k-1;i>=0;i--) cout<<arry1[i];
return 0;
}
原文：http://blog.csdn.net/zju_ziqin/article/details/19765417