【PAT】【Advanced Level】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:

Yes2469135798

原题链接：

https://www.patest.cn/contests/pat-a-practise/1023

思路：

按要求计算统计即可

CODE：

#include<iostream>#include<cstring>#include<string>using namespace std;int main(){    string a;    cin>>a;    int c[a.length()+1];    int flag1[10];    int flag2[10];    memset(flag1,0,sizeof(flag1));    memset(flag2,0,sizeof(flag2));    memset(c,0,sizeof(c));    for (int i=a.length()-1;i>=0;i--)    {        flag1[a[i]-'0']=1;        c[i+1]+=(a[i]-'0')*2;        c[i]=c[i+1]/10;        c[i+1]%=10;    }    if (c[0]!=0) flag2[c[0]]=1;    for (int i=1;i<=a.length();i++)        flag2[c[i]]=1;    int re=0;    for (int i=0;i<10;i++)        if (flag1[i]!=flag2[i])    {        re=1;        break;    }    if (re==0) cout<<"Yes"<<endl;    else        cout<<"No"<<endl;    if (c[0]!=0)cout<<c[0];    for (int i=1;i<=a.length();i++)        cout<<c[i];    return 0;}