[CCPC2020绵阳K] Knowledge is Power - 构造,找规律

[CCPC2020绵阳K] Knowledge is Power - 构造,找规律

Description

给你一个数字x，把x至少拆分为两个数的和，并且这些数字互质，问所有拆分方案中，在每个方案中最大值减去最小值的最小值是多少

Solution

设 \(x=4y+r\)，对 \(r\) 讨论

如果 \(r=0\)，拆成 \(x/2+1,x/2-1\)

如果 \(r=1,3\)，拆成 \([x/2],[x/2]+1\)

剩下的情况比较麻烦，但是经过一些尝试后，发现答案其实不可能超过 \(4\)

那么枚举几种拆分方案就可以

下面的代码中是暴力对拆成两个，拆成三个的方法进行尝试的

#include <bits/stdc++.h>
using namespace std;

#define int long long
int caseid = 0;

void solve()
{
    ++caseid;
    int n;
    cin >> n;
    int r = n % 4;
    cout << "Case #" << caseid << ": ";
    if (n == 6)
    {
        cout << -1 << endl;
    }
    else if (r == 0)
    {
        cout << 2 << endl;
    }
    else if (r == 1 || r == 3)
    {
        cout << 1 << endl;
    }
    else
    {
        int m = n / 4;
        int c = 2 * m / 3;
        int ans = 1e18;
        for (int i = n / 2 - 4; i <= n / 2; i++)
        {
            int j = n - i;
            if (i <= 1 || j <= 1 || i % 2 == 0)
                continue;
            if (__gcd(i, j) == 1)
            {
                ans = min(ans, j - i);
            }
        }
        for (int i = c - 10; i <= c + 10; i++)
        {
            int j = m - i;
            if (i <= 0 || j < 0)
                continue;
            int a = 2 * i - 1;
            int b = 2 * i + 1;
            int c = n - a - b;
            if (a <= 1)
                continue;
            if (c <= 1)
                continue;
            int tmp = abs(a - b);
            tmp = max(tmp, abs(a - c));
            tmp = max(tmp, abs(b - c));
            if (tmp > ans)
                continue;
            if (__gcd(a, c) == 1 && __gcd(b, c) == 1)
                ans = min(ans, tmp);
        }
        cout << ans << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);

    int t;
    cin >> t;
    while (t--)
        solve();
}