Codeforces Round 901 (Div. 2)

合集 - codeforces/atcoder补题(18)

1.Atcoder Beginner Contest 3122023-07-302.Educational Codeforces Round 152 (Rated for Div. 2)2023-07-283.Codeforces Round 889 (Div. 2)2023-07-314.Codeforces Round 888 (Div. 3)2023-07-275.Codeforces Round 847 (Div. 3)2023-08-086.Codeforces Round 890 (Div. 2)2023-08-107.Codeforces Round 892 (Div. 2)2023-08-138.Codeforces Round 893 (Div. 2)2023-08-179.Educational Codeforces Round 1532023-08-1910.Codeforces Round 891 (Div. 3)2023-08-1011.Codeforces Round 894 (Div. 3)2023-08-2912.Educational Codeforces Round 154 (Rated for Div. 2)2023-09-0713.Codeforces Round 896 (Div. 2)2023-09-1314.Codeforces Round 855 (Div. 3)2023-09-1815.codeforces round 895 (div. 3)2023-09-1116.Codeforces Round 900 (Div. 3)2023-09-28

17.Codeforces Round 901 (Div. 2)2024-09-06

18.Codeforces Round 903 (Div. 3)2023-10-15

收起

C

题意：

给你n个青苹果要分给m个人，不够分的时候可以平均切一刀，问最少且多少刀可以让每个人得到的苹果一样多

思路：

贪心的想肯定是先把一整个苹果分给大家，这样就不用切了，所以 n = n % m, 多出来的苹果再切完分
最少要切成lcm(n, m)份才能均分，由于每次都只能对半切，所以lcm(n, m) / n一定是$2^x$个，
平均下来一个人lcm(n, m) / m片，因为每切一刀果子片数+1，那么m个人都有 刀数 * m片

inline void solve() 
{
	int n, m; cin >> n >> m;
	n = n % m;

	if (n == 0)
	{
		cout << 0 << endl;
		return;
	}

	LL lcm = 1LL * n * m / gcd(n, m); // 最少要有lcm片才能均分给m个人
	LL a = lcm / n; // 每个GreenApple应该切成的片数
	LL b = lcm / m; // 每个人应该得到的片数

	if ((a & -a) != a)
	{
		cout << -1 << endl;
		return;
	}

	int cnt = 0;
	while (b)
	{
		b = b - (b & -b);
		cnt++;
	}

	cout << 1LL * cnt * m - n << endl; 

}