1023 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.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

tip:该题既然只是两倍，那就当作大数相加的逻辑进行运算，然后正常写一套即可

 

#include<bits/stdc++.h>
using namespace std;
int main(){
    char a[25],b[25],c[25];
    int cnta[260]={0},cntc[260]={0};
    bool flag=true;
    cin>>a;
    strcpy(b,a);
    memset(c,0,sizeof(c));
    for (int i=strlen(a)-1;i>=0;--i){
        c[i]=(a[i]-'0'+b[i]-'0')%10+'0';
        if (i-1>=0){
            b[i-1]+=(a[i]-'0'+b[i]-'0')/10;
        }
        cnta[a[i]]++;
        cntc[c[i]]++;
    }
    if (a[0]<'5'){
        for (int i=0;i<=256;++i){
            if (cnta[i]!=cntc[i]){
                flag=false;
            }
        }
    }
    else
        flag=false;
    
    if (flag)
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
    if (a[0]>='5'){
        cout<<'1';
    }
    cout<<c<<endl;
}