连求两题

沐咕 @ 2024-11-08 21:34:30

#include <bits/stdc++.h>
using namespace std;
long long a, b;
set<long long> s;
int main()
{
    cin >> a >> b;
    s.insert(a);
    s.insert(b);
    long long x = a, y = b;
    while (true)
    {
        if (s.find(abs(y - x)) != s.end())
            break;
        s.insert(abs(y - x));
        x = y;
        y = abs(y - x);
    }
    cout << s.size() << endl;
    return 0;
}

by MLE_Automaton @ 2024-11-08 21:45:21

@沐咕 结束条件是S_i=0吧

by 沐咕 @ 2024-11-08 21:45:28

球球了各位大佬，给个代码吧qwq

by Argon_Cube @ 2024-11-08 21:45:42

每次需要插入 set 的是 (S_{i-1},S_{i})。

by DDD_et @ 2024-11-08 21:46:00

@沐咕

纯 WA 啊，那行，等我做一下

by 沐咕 @ 2024-11-08 21:46:10

@Argon_Cube 所以咋改，我要代码qwq

by ran_qwq @ 2024-11-08 21:46:13

@沐咕 出现重复的数不能保证以后的数都是重复的

如10->2->8->6->2->4

注意这个4在出现两次的2之后出现，但它是一个新的数

by __Louis__ @ 2024-11-08 21:46:24

#include <bits/stdc++.h>
using namespace std;
long long a, b;
set<long long> s;
int main()
{
    cin >> a >> b;
    s.insert(a);
    s.insert(b);
    long long x = a, y = b;
    while (true)
    {
        if (!s.count(x)&&!s.count(y))
            break;
        s.insert(abs(y - x));
        long long tx = x;
        x = y;
        y = abs(y - tx);
    }
    cout << s.size() << endl;
    return 0;
}

by 沐咕 @ 2024-11-08 21:46:24

@DDD_et OK

by __Louis__ @ 2024-11-08 21:47:00

@沐咕

#include <bits/stdc++.h>
using namespace std;
long long a, b;
set<long long> s;
int main()
{
    cin >> a >> b;
    s.insert(a);
    s.insert(b);
    long long x = a, y = b;
    while (true)
    {
        if (!s.count(x)&&!s.count(y))
            break;
        s.insert(abs(y - x));
        long long tx = x;
        x = y;
        y = abs(y - tx);
    }
    cout << s.size() << endl;
    return 0;
}

by __Louis__ @ 2024-11-08 21:47:41

等等，有 bug。