pat1023. 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

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提交代码

 

#include<set>
#include<map>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
using namespace std;
int num[25],dight[10];
int main(){
    //freopen("D:\\input.txt","r",stdin);
    string s;
    cin>>s;
    int i;
    for(i=0;i<s.length();i++){
        num[i]=s[s.length()-1-i]-'0';
        
        //cout<<num[i]<<endl;
        
        dight[num[i]]++;
        num[i]*=2;
        
        //cout<<num[i]<<endl;
    }
    int k=0;
    for(i=0;i<s.length();i++){
        int sum=num[i]+k;
        num[i]=sum%10;
        k=sum/10;
    }
    
    if(k){
        num[i++]=k;
    }
    int len=--i;
    for(;i>=0;i--){
        dight[num[i]]--;
    }
    for(i=0;i<=9;i++){
        if(dight[i]){
            break;
        }
    }
    
    if(i==10){
        printf("Yes\n");
    }
    else{
        printf("No\n");
    }
    for(i=len;i>=0;i--){
        cout<<num[i];
    }
    cout<<endl;
    return 0;
}