SGU113 Nearly prime numbers

SGU113 Nearly prime numbers

题目大意

算法思路

时间复杂度: O(V‾‾√)O(\sqrt V)

代码

/**
* Copyright (c) 2015 Authors. All rights reserved.
*
* FileName: 113.cpp
* Author: Beiyu Li <sysulby@gmail.com>
* Date: 2015-05-22
*/
#include <bits/stdc++.h>

using namespace std;

#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)

typedef long long LL;
typedef pair<int, int> Pii;

const int inf = 0x3f3f3f3f;
const LL infLL = 0x3f3f3f3f3f3f3f3fLL;

const int maxv = 100000 + 5;

bool isp[maxv];
int prime[maxv], sz;
void gen_prime()
{
sz = 0;
memset(isp, true, sizeof(isp));
isp[0] = isp[1] = false;
for (int i = 2; i < maxv; ++i) {
if (isp[i]) prime[sz++] = i;
for (int j = 0, t; j < sz && (t = prime[j] * i) < maxv; ++j) {
isp[t] = false;
if (i % prime[j] == 0) break;
}
}
}

int main()
{
gen_prime();
int n, v;
scanf("%d", &n);
rep(i,n) {
int t = 0;
scanf("%d", &v);
rep(j,sz) {
int p = prime[j];
if (p * p > v) break;
while (v % p == 0) ++t, v /= p;
}
if (v > 1) ++t;
puts(t == 2? "Yes": "No");
}

return 0;
}