PAT 1023 Have Fun with Numbers（高精度乘法）

1023 Have Fun with Numbers

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

2022.11.1（可能之前忘记写了） 

代码：两次代码思路都是一样的

//思路：先将原始出现的数字存储下来，然后将数字乘2（高精度乘法），最后判断2倍之后的数字是否
//出现了和原始数字不同的数字
#include 
#include 
using namespace std;
 
vector<int> b(10);
vector<int> mul(vector<int> &a,int b){
    vector<int> c;
    int t=0;
    
    for(int i=0;isize() || t;i++){
        t+=a[i]*b;
        c.push_back(t%10);
        t/=10;
    }
    //除前导0的情况是，0000*1234，结果为0000，那么需要除去前导0，变成0
    while(c.size()>1 && c.back()==0)    c.pop_back();
    return c;
}
int main(){
    string s;
    vector<int> a;
    cin >> s;
    
    for(auto c:s)   b[c-'0']++;//记录出现的数字
    for(int i=s.size()-1;i>=0;i--)  a.push_back(s[i]-'0');
    auto c=mul(a,2);
    bool sign=false;
    for(auto q:c){
        b[q]--;
        if(b[q]<0){
            sign=true;
            break;
        }
    }
    if(!sign)    puts("Yes");
    else puts("No");
    for(int i=c.size()-1;i>=0;i--)  printf("%d",c[i]);
    return 0;
}

 好好学习，天天向上！

我要考研！