lightOJ1341(唯一分解定理)

前面学的好多知识都忘了，现在决定复习一下了

问有几种边长为整数的矩形面积等于a，且矩形的短边不小于b

用唯一分解定理求出a的因子对数，在暴力求出1-b内的a的因子个数，相减一下就好了

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>

using namespace std;
const int mx = 1e6+7;
bool vis[mx];
int prime[mx];
int cnt; 

void sieve(long long n) {
	int m = (int)sqrt(n+0.5);
	memset(vis, 0, sizeof(vis));
	for (int i = 2; i <= m; i++) if (!vis[i])
		for (int j = i*i; j <= n; j+=i) vis[j] = 1;
}

int gen_primes(int n) {
	sieve(n);
	int c = 0;
	for (int i = 1; i <= n; i++) if (!vis[i])
		prime[c++] = i;
	return c;
}

int main()
{
	cnt = gen_primes(mx-1);
	int T, kase = 0;
	scanf("%d",&T);
	
	while (T--)
	{
		long long a, b;
		scanf("%lld%lld",&a,&b);
		long long ans = 1;
		if (b >= sqrt(a))
		{
            printf("Case %d: %d\n",++kase,0);
            continue;
        }
		else
		{
			long long tmp = a;
			for (int i = 1; i < cnt && 1LL*prime[i]*prime[i] <= a; i++)
			{
				if (a % prime[i] == 0)
				{
					int num = 0;
					while (a % prime[i] == 0)
					{
						num++;
						a /= prime[i];
					}
					ans *= (num + 1);
				}
			}

			if (a > 1) ans*= 2;//最后分解完可能剩一个质数分解不掉，所以这里要乘以2 
			ans /= 2;
			for (int i = 1; i < b; i++)
				if (tmp % i == 0) ans--;
		}
		printf("Case %d: %lld\n",++kase,ans);
	}
	return 0;
}