Have Fun with Numbers

自测-4 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

 

 

 

 

#include <iostream>
#include <vector>
using namespace std;
int main(){
    string s1;
    cin >> s1;//将数字按字符串的格式读入 
    vector<int> myVec; 
    vector<int> myVec1;
    for(int i = 0; i < s1.size(); i++){
        myVec.push_back(s1[i] - '0');//依次将字符转化为数字存入vector中 
        myVec1.push_back(2 * (s1[i] - '0'));//依次将字符转化为数字的二倍存入vector中 
    }
    //这里请注意:上面将每一位扩大为原来数字的二倍可能会大于9
    //所以要将每一位进位处理 
    //从最后一位开始处理 
    for(int i = myVec1.size() - 1; i >= 0; i--){
        if(i != 0){
            myVec1[i - 1] = myVec1[i] / 10 + myVec1[i - 1];
        
            myVec1[i] %= 10;
                
        }else{
            break;
        }    
    
    } 
    //分别开辟俩个大小为10的vector
    //用以记录0-9各数字的个数 
    vector<int> myVec2(10, 0);
    
    vector<int> myVec3(10, 0);
    //如果上面的进位处理会使总位数增加，也就是第一位会大于9，则输出"No" ,并遍历结果 
    if(myVec1[0] > 9){
        cout << "No" << endl;
        for(int i = 0; i < myVec1.size(); i++){
            cout << myVec1[i];
        }
        cout << endl;
        return 0;
    }
    //统计0-9数字各自的个数 
    for(int i = 0; i < myVec.size(); i++){
        myVec2[myVec[i]]++;
        myVec3[myVec1[i]]++;
    }

    int n;
    for(n = 0; n < 10; n++){
        if(myVec2[n] != myVec3[n]){    //如果对应数字个数不相同，则跳出循环 
            break;
        }
    }

    if(n == 10){//如果n==10则说明没有异常输出 
        cout << "Yes" << endl;
        for(int i = 0; i < myVec1.size(); i++){
            cout << myVec1[i];
        }
        cout << endl;
    }else{
        cout << "No" << endl;
        for(int i = 0; i < myVec1.size(); i++){
            cout << myVec1[i];
        }
        cout << endl;
    }
    return 0;
}