摘要
Title: A1023 Have Fun with Numbers
Tag: 高精度
Memory Limit: 64 MB
Time Limit: 1000 ms

A1023 Have Fun with Numbers

题意

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.

思路

如果用Python3，则可以用Counter直接模拟即可
如果用C++

int 是 $[-2*10^9 , 2*10^9]$
long long 是 $[-9*10^{18} , 9*10^{18}]$
这个题有20位，会超，所以要开高精度
高精度需要从低位开始处理，所以字符串放进vector时要倒着放，输出时也要倒序输出

高精加

vector<int> add(vector<int> &A, vector<int> &B) // 加上引用可以提高效率，防止函数将vector全拷贝过来
{
    if (A.size() < B.size()) return add(B, A); // 调整A的size最大

    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++ )
    {
        t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }

    if (t) C.push_back(t);
    return C;
}

高精乘

vector<int> mul(vector<int> &A, int b)
{
    vector<int> C;

    int t = 0;
    for (int i = 0; i < A.size(); i ++ )
    {
        t += A[i] * b;
        C.push_back(t % 10);
        t /= 10;
    }

    if(t) C.push_back(t);

    while (C.size() > 1 && C.back() == 0) C.pop_back(); // 去掉前导0
    return C;
}

代码

python3

'''
Author: NEFU AB-IN
Date: 2023-01-06 10:02:20
FilePath: \GPLT\A1023\A1023.py
LastEditTime: 2023-01-06 10:39:00
'''
from collections import Counter

n = input()
n2 = str(int(n) * 2)

d1 = Counter(n)
d2 = Counter(n2)

if d1 == d2:
    print("Yes")
else:
    print("No")

print(n2)

C++

/*
* @Author: NEFU AB-IN
* @Date: 2023-01-06 18:00:41
* @FilePath: \GPLT\A1023\A1023.cpp
* @LastEditTime: 2023-01-06 18:09:24
*/
#include <bits/stdc++.h>
using namespace std;
#define N n + 100
#define int long long
#define SZ(X) ((int)(X).size())
#define IOS                                                                                                            \
    ios::sync_with_stdio(false);                                                                                       \
    cin.tie(nullptr);                                                                                                  \
    cout.tie(nullptr)
#define DEBUG(X) cout << #X << ": " << X << endl
typedef pair<int, int> PII;

// #undef N
// const int N = 1e5 + 10;

// #undef int

vector<int> add(vector<int> A, vector<int> B)
{
    vector<int> C;
    int t = 0;

    for (int i = 0; i < SZ(A); i++)
    {
        int s = A[i] + B[i] + t;
        C.push_back(s % 10);
        t = s / 10;
    }

    if (t)
        C.push_back(t);
    return C;
}

signed main()
{
    IOS;
    string s;
    cin >> s;

    vector<int> a;
    for (int i = SZ(s) - 1; i >= 0; --i)
        a.push_back(s[i] - '0');

    vector<int> b = add(a, a);
    vector<int> c = b;
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());

    cout << (a == b ? "Yes" : "No") << '\n';
    for (int i = SZ(c) - 1; i >= 0; --i)
        cout << c[i];
    return 0;
}

A1023 Have Fun with Numbers

题意

思路

代码