PAT A1023 Have Fun with Numbers(20)[趣味数字，高精度]

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

题意

一个大整数倍乘之后的结果与源数的数字是否相同

思路1

1. 模拟大整数加法，自身相加，得到结果
2. 将 vector 排序，若结果相同，表示可以重组，否则不行

代码1

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
    vector<int> a, ans;
    string str;
    cin >> str;
    for(int i = str.size() - 1; i >= 0; i--)
        a.push_back(str[i] - '0');
    int t = 0; // 进位 
    for(int i = 0; i < a.size(); i++)
    {
        t += a[i] * 2;
        ans.push_back(t % 10);
        t /= 10;
    }
    if(t)
        ans.push_back(t);
    sort(a.begin(), a.end());
    vector<int> b(ans); 
    sort(b.begin(), b.end());
    if(a == b)
        puts("Yes");
    else
        puts("No");
    for(int i = ans.size() - 1; i >= 0; i--)
        printf("%d", ans[i]);
    puts("");
    return 0;
}