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Codeforces Round #209 (Div. 2) / 359D Pair of Numbers (一点点想法)


http://codeforces.com/problemset/problem/359/D


/*124ms,2300KB*/

#include<cstdio>

int a[300005], w[300005];

int main()
{
	int n, i, l, r, cnt = 0, maxd = 0;
	scanf("%d", &n);
	for (i = 0; i < n; i++) scanf("%d", &a[i]);
	for (i = 0; i < n;)
	{
		l = r = i;
		while (l && a[l - 1] % a[i] == 0) l--; ///左移
		while (r < n - 1 && a[r + 1] % a[i] == 0) r++; ///右移
		i = r + 1; /// 关键一步,保证算法是O(n)的,这样做是利用“连续性”及这些数已经能被a[i]整除
		r -= l; ///现在r相当于r-l
		if (r > maxd) cnt = 0, maxd = r;
		if (r == maxd) w[cnt++] = l + 1;
	}
	printf("%d %d\n", cnt, maxd);
	for (i = 0; i < cnt; ++i) printf("%d ", w[i]); ///多打一个空格没问题
	return 0;
}


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