[原创]“玲珑杯”ACM比赛 Round #9 A – Check-in Problem [因子个数]【数论】

2017-02-10 21:31:20 Tabris_ 阅读数：440

博客爬取于2020-06-14 22:41:42
以下为正文

版权声明：本文为Tabris原创文章，未经博主允许不得私自转载。
https://blog.csdn.net/qq_33184171/article/details/54974236

题目连接:http://www.ifrog.cc/acm/problem/1084

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A – Check-in Problem
Time Limit：5s Memory Limit：128MByte

Submissions：921 Solved：55

DESCRIPTION

A positive integer x is called p-bizarre number if the number of the divisors of x is p exactly.
Your task is testing whether the given positive integer n is a p-bizarre number or not.

INPUT

The first line contains a positive integer T, which represents there are T test cases.
The following is test cases. For each test case:
The only one line contains a positive integer n and an odd prime p.
1≤T≤10^5,1≤n≤10^18,2< p≤10^9

OUTPUT

For each test case, output in one line, print “YES” (without quote) if n is a p-bizarre number, print “NO” (without quote) otherwise.

SAMPLE INPUT

3
9 3
971528476274196481 7
150094635296999121 37

SAMPLE OUTPUT
YES
NO
YES
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题目大意:
就是问你n的因子个数是不是p个

解题思路:
对于一个素数n的因子个数 我们可以对n做算术基本定理展开
n=p1a1×p2a2×p3a3×...×prarn = p_1^{a_1}\times p_2^{a_2}\times p_3^{a_3}\times ...\times p_r^{a_r}

那么数的因子个数就是∑i=1r(a1+1)×(a2+1)×(a3+1)×...×(an+1)\sum_{i = 1}^{r}(a_1+1)\times (a_2+1)\times (a_3+1)\times ...\times (a_n+1)

input里面又说
The only one line contains a positive integer n and an odd prime p.

那么对于p是素数的情况 只能说明n的质因子只有一种,

因为上述,所以我想到以对1e6(因为题目说p最小是3,n最大是10^18,所以1e6就够了)内的素数筛法取一遍,然后二分寻找答案即可注意会爆LL ,但是无论怎么控制溢出,最后代码写成了这样但是还是WA…心塞…

献上官方题解

注意到pp 是质数，只有当nn 是质数的 $p−1 $次幂时，nn 的约数才可能恰好有pp 个，所以判定一个正整数nn 是p−p-奇异数，只需检验p−1n^{p-1}\sqrt {n} 是整数，且p−1n^{p-1}\sqrt {n} 是质数。预处理109\sqrt {10^9} 以内的素数（共34013401 个），进行开根和判断素数即可，时间复杂度$ O\left(\dfrac {\sqrt n}{ \ln⁡ n}\right) $。
事实上p>3p>3 的情况很少有解，直接预处理所有有解的情况即可，可以防止写出有问题的开根，而p=3p=3 的判断素数也可以用 Miller-Rabin 算法判定（需要O(1)O(1) 的模乘法）。

改了2个小时的溢出，最后都没签到。。。。。。

献上标程一枚
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# include <cmath>
# include <stdio.h>
# include <cassert>
# include <algorithm>
typedef long long LL;
const int maxn = 31623, maxm = 17, maxp = 61, maxt = 100001, maxv2 = (int)1e9;
const LL maxv = (LL)1e18;
int tot, pr[maxn], d[maxn], sz[maxm];
LL pp[maxm][maxn];
bool isprime(int x)
{
	if(x < 2)    return 0;
	if(x < maxn) return d[x] == x;
	for(int i = 0; i < tot && pr[i] * pr[i] <= x; ++i)
		if(x % pr[i] == 0)	return 0;
	return 1;
}
int main()
{
	for(int i = 2; i < maxn; ++i)
	{
		if(!d[i])
			pr[tot++] = d[i] = i;
		for(int j = 0, k; (k = i * pr[j]) < maxn; ++j)
		{
			d[k] = pr[j];
			if(d[i] == pr[j])break;
		}
	}

	for(int i = 2; i < maxm; ++i)
		for(int j = 0; j < tot; ++j)
		{
			int rem = pr[i] - 1;
			LL val = 1, lim = maxv / pr[j];
			for( ; rem && val <= lim; --rem, val *= pr[j]);
			if(rem)   	break;
			pp[i][sz[i]++] = val;
		}

	int t;
	LL n, p;
	assert(scanf("%d", &t) == 1
	&& 1 <= t && t < maxt);
	while(t--)
	{
		assert(scanf("%lld%lld", &n, &p) == 2
		&& 1 <= n && n <= maxv
		&& (p & 1) && p <= maxv2 && isprime(p));
		if(p >= maxp || d[p] != p)
		{
			puts("NO");
			continue;
		}
		if(p == 3)
		{
			LL val = (LL)sqrt(n);
			for( ; val * val > n; --val);
			for( ; (val + 1) * (val + 1) <= n; ++val);
			puts(val * val == n && isprime(val) ? "YES" : "NO");
			continue;
		}
		for(int i = 2; i < maxm; ++i)
			if(pr[i] == p)
			{
				puts(*std::lower_bound(pp[i], pp[i] + sz[i], n) == n ? "YES" : "NO");
				break;
			}
	}
	return 0;
}