LOJ 1341 Aladdin and the Flying Carpet（质因子分解）

题目链接：http://lightoj.com/volume_showproblem.php?problem=1341

题意：给两个数a，b，求满足c * d = a且c>=b且d>=b的c, d二元组对数，(c, d)和(d, c)属于同一种情况。

思路：根据唯一分解定理，先将a唯一分解，则a的所有正约数的个数为num = (1 + a1) * (1 + a2) *...* (1 + ai)，这里的ai是素因子的指数。现在我们知道了a的因子个数为num，假设因子从小到大排列为

X1,X2,...,Xnum 因为是求二元组，所以num先除以2，然后减去那些比b小的因子即可（比b小的一定和比较大的组合，所以不用考虑很大的情况）。

code：

#include <cstdio>
#include <cstring>
using namespace std;

typedef long long LL;
const int MAXN = 1000000;

int prime[MAXN + 1];

void getPrime()
{
    memset(prime, 0, sizeof(prime));
    for (int i = 2; i <= MAXN; ++i) {
        if (!prime[i]) prime[++prime[0]] = i;
        for (int j = 1; j <= prime[0] && prime[j] <= MAXN / i; ++j) {
            prime[prime[j] * i] = 1;
            if (i % prime[j] == 0) break;
        }
    }
}

LL factor[1000][2];
int fatCnt;

int getFactors(LL x)
{
    fatCnt = 0;
    LL tmp = x;
    for (int i = 1; i <= prime[0] && prime[i] * prime[i] <= tmp; ++i) {
        factor[fatCnt][1] = 0;
        if (tmp % prime[i] == 0) {
            factor[fatCnt][0] = prime[i];
            while (tmp % prime[i] == 0) {
                ++factor[fatCnt][1];
                tmp /= prime[i];
            }
            ++fatCnt;
        }
    }
    if (tmp != 1) {
        factor[fatCnt][0] = tmp;
        factor[fatCnt++][1] = 1;
    }
    LL ret = 1;
    for (int i = 0; i < fatCnt; ++i) {
        ret *= (1L + factor[i][1]);
    }
    return ret;
}

int main()
{
    getPrime();
    int nCase;
    scanf("%d", &nCase);
    for (int cnt = 1; cnt <= nCase; ++cnt) {
        LL a, b;
        scanf("%lld %lld", &a, &b);
        if (b * b > a) {
            printf("Case %d: 0
", cnt);
            continue;
        }
        LL num = getFactors(a);
        num /= 2;
        for (int i = 1; i < b; ++i) {
            if (a % i == 0) --num;
        }
        printf("Case %d: %lld
", cnt, num);
    }
    return 0;
}