ACM:每日一题 abc336 C题

C - Even Digits Editorial

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score: 300300 points

Problem Statement

A non-negative integer n is called a good integer when it satisfies the following condition:

• All digits in the decimal notation of n are even numbers (0, 2, 4, 6, and 8).

For example, 0, 68, and 2024 are good integers.

You are given an integer N. Find the N-th smallest good integer.

Constraints

• 1≤N≤10^12
• N is an integer.

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Input

The input is given from Standard Input in the following format:

N

Output

Print the N-th smallest good integer.

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Sample Input 1

Copy

8

Sample Output 1

Copy

24

The good integers in ascending order are 0,2,4,6,8,20,22,24,26,28,…0,2,4,6,8,20,22,24,26,28,….
The eighth smallest is 2424, which should be printed.

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Sample Input 2

133

Sample Output 2

2024

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Sample Input 3

31415926535

Sample Output 3

2006628868244228

初次编写代码测试：

#include
using namespace std;
#define int long long 
int n;
int fun(int n){
    int p,i=0;
    if(n==1)return i;
    for(p=2,i=0;p<=n;p++){
        i+=2;
        int q=i,cnt=0;
        
        while(q){
        int t=q%10;
        if(t%2!=0){i+=pow(10,cnt);break;}
        q/=10;
        cnt++;
       }
    }
    return i;
}
signed main(){
    cin>>n;
    cout<

改进（10^12想都不要想）：

要精用vector，数组一般都会超时的啦。

#include 
using namespace std;
#define int long long
vector a;//类似高精度
int n;
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cin >> n;
    n-=1;//因为第一位是0嘛
    while(n > 0){//2*5=10进位了
        a.push_back(n%5);
        n /= 5;
    }
    if(a.empty()) a.push_back(0);
    for(int i = a.size()-1; i > -1; i--)//迭代器，也不算吧，vector也是数组
        cout << 2*a[i];
}

也算是运用了算法吧这个题目，高精度还是很重要的（上一次选拔赛就是高精度差一点就提交了ac了）。

这篇不算入每日一题的范畴。是acm作业吧。