1023 Have Fun with Numbers

1023 Have Fun with Numbers
分数 20

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作者 CHEN, Yue
单位 浙江大学
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

1.分析

        1.20位数字，用字符串进行运算。

        2.数字标记出现数字的次数，a中出现加一，b中出现减一。最后判断是否为0即可。

2.代码

#include<iostream>
#include<vector>
using namespace std;
string a,b;
vector<int> v;
int idx[12],t;
bool check(){          //判断for(int i=0;i<10;i++){if(idx[i]!=0) return false;}return true;
}
int main(){cin>>a;for(int i=0;i<a.size();i++){     //标记aidx[a[i]-'0']++;}for(int i=a.size()-1;~i;i--){    //计数bt+=(a[i]-'0')*2;            //进位加法idx[t%10]--;v.push_back(t%10);        //存储t/=10;}if(t!=0) idx[t]--,v.push_back(t);if(check()) cout<<"Yes"<<endl;else cout<<"No"<<endl;for(int i=v.size()-1;i>=0;i--){   //输出cout<<v[i];}return 0;
}