1085 Perfect Sequence

解题代码

#include
#include
using namespace std;
const int nmax = 100010;
int a[nmax], n, p, cnt;
int main() {
	scanf("%d%d", &n, &p);
	for (int i = 0; i < n; i++) scanf("%d", &a[i]);
	sort(a, a + n);
	for (int i = 0; i < n; i++) {
		int j = upper_bound(a + i + 1, a + n, (long long)p*a[i]) - a;
		if (j - i > cnt) cnt = j - i;
	}
	printf("%d", cnt);
	return 0;
}

测试结果

问题整理

1.二分法。