PAT PAT 1023. Have Fun with Numbers (20)

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 withk 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:

Yes2469135798题意：给你一个k位的整数，然后将它乘以2，得到结果sum，如果结果sum的组成数字出现的次数和给出的数相同，那么我们就输出Yes，反之，输出No，
然后再把sum输出。#include <iostream>#include <cstring>#include <cstdio>using namespace std;int main(){    char s[30],c[30],d[30];    int len,i,j,cnt,cnt1[15],cnt2[15],x,flag;    while(cin>>s)    {        len=strlen(s);        cnt=0;        for(i=len-1;i>=0;i--)        {            x=(s[i]-'0')*2;            c[i]=x%10+cnt+'0';//把进位后的数组存起来            cnt=x/10;//进位        }        if(cnt==1)//如果在最后一次运算都有进位，那么肯定是错的，因为相同的数字出现了多次        {            cout<<"No"<<endl;            cout<<"1";//因为乘以2，所以进位为1            for(i=0;i<len;i++)            {                cout<<c[i];            }            cout<<endl;            return 0;        }            memset(cnt1,0,sizeof(cnt1));            memset(cnt2,0,sizeof(cnt2));            for(j=0;j<len;j++)            {                cnt1[s[j]-'0']++;//把出现的数字的次数加起来                cnt2[c[j]-'0']++;            }            flag=1;            for(j=0;j<10;j++)            {                if(cnt1[j]!=cnt2[j])//如果两个数组中数字出现的次数不同，那么肯定是错的                {                    flag=0;                    break;                }            }            if(flag)            {                cout<<"Yes"<<endl;            }else{                cout<<"No"<<endl;            }            for(i=0;i<len;i++)            {                cout<<c[i];            }            cout<<endl;    }    return 0;}