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 21:38:58

@沐咕 这个应该是出现0才循环吧

by 沐咕 @ 2024-11-08 21:39:36

玄关

• 我没滥用标题行！
• 我真没滥用标题行，不是我！
• 都说了我没滥用标题行！

by DDD_et @ 2024-11-08 21:39:45

@MLE_Automaton

有道理，出现 0 或者像我刚才说的那样都会循环

by ran_qwq @ 2024-11-08 21:40:18

@ran_qwq *x已经被赋值成y了

by DDD_et @ 2024-11-08 21:40:37

@沐咕

当 x = y 后 abs (y - x) 难道不等于 0 吗……

by 沐咕 @ 2024-11-08 21:40:52

@ran_qwq

这样也不对，还是 10 分（

#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));
        long long xx = x, yy = y;
        x = yy;
        y = abs(yy - xx);
    }
    cout << s.size() << endl;
    return 0;
}

by __Louis__ @ 2024-11-08 21:41:10

这题不应该是什么 gcd 之类的题吗，用set大概率会T

by DDD_et @ 2024-11-08 21:41:25

改成这样？

while (true)
    {
        if (s.find(abs(y - x)) != s.end())
            break;
        s.insert(abs(y - x));
        long long tx = x;
        x = y;
        y = abs(y - tx);
    }

by 沐咕 @ 2024-11-08 21:41:31

@DDD_et @ran_qwq 所以能给个代码吗，不知道咋改，知道问题出在哪qwq

by DDD_et @ 2024-11-08 21:41:46

用来替换原来的 while 循环