洛谷1072 hankson的趣味题数论乱搞非标准解法

最新推荐文章于 2025-08-04 17:49:12 发布

原创最新推荐文章于 2025-08-04 17:49:12 发布 · 290 阅读

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本文针对一道涉及数学原理的算法题目提供了详细的解题思路与代码实现。通过对题目要求的逐步解析，利用数学公式推导出高效的求解方法，并通过具体代码实现了优化后的算法。

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题解：

看起来这是一道数学题，，，那么，，先打个暴力放松一下心情。。

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


int n;
long long anstot;
long long a0, a1, b0, b1;

long long gcd(long long x, long long y) {
    if (y == 0) return (x);
    return (gcd(y, x % y));
}
int main () {
    freopen("son.in", "r", stdin);
    freopen("son.out", "w", stdout);
    scanf("%d", &n);
    while (n--) {
        anstot = 0;
        scanf("%lld %lld %lld %lld", &a0, &a1, &b0, &b1);
        for (long long x = a1; x <= b1; x++) {
            if (gcd(x, a0) != a1) continue;
            long long tx = gcd(x, b0);
            if (x * b0 / tx != b1) continue;
            anstot++;
        }
        printf("%lld\n", anstot);
    }

    return 0;
}

找一张纸简单的推导一下看看有没有啥有用的发现：
$∵ l c m (x, b 0) = b 1$ $\because lcm(x, b_0) = b_1$
$∴ x * b 0 / g c d (x, b 0) = b 1$ $\therefore x*b_0/gcd(x,b_0)=b_1$
$∴ g c d (x, b 0) = x * b 0 / b 1$ $\therefore gcd(x,b_0) = x*b_0/b_1$
$∴ g c d (b 1 / b 0, b 1 / x) = 1$ $\therefore gcd(b_1/b_0, b_1/x) =1$
显然x是 $b_1$ 的因数，那么直接从1到 $\sqrt(n)$ 枚举x判断是否成立即可

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

int n;
long long anstot;
long long a0, a1, b0, b1;

long long gcd(long long x, long long y) {
    if (y == 0) return (x);
    return (gcd(y, x % y));
}
void check(long long x) {
    if (gcd(x, a0) != a1) return;
    long long cur = gcd(x, b0);
    if (x * b0 / cur != b1) return;
    anstot++;
}
int main () {
    freopen("son.in", "r", stdin);
    freopen("son.out", "w", stdout);
    scanf("%d", &n);
    while (n--) {
        anstot = 0;
        scanf("%lld %lld %lld %lld", &a0, &a1, &b0, &b1);
        long long c = sqrt(b1);
        for (long long x = 1; x <= c; x++) {
            if (b1 % x != 0) continue;
            check(x);
            if (x * x == b1) continue;
            check(b1 / x);
        }
        printf("%lld\n", anstot);
    }

    return 0;
}

90分了，，，再想想，把long long改成int，check的时候压一下常数。。

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

int n;
long long anstot;
int a0, a1, b0, b1;

int gcd(int x, int y) {
    if (y == 0) return (x);
    return (gcd(y, x % y));
}
void check(int x) {
    if (x % a1) return;
    if (gcd(x/a1, a0/a1) == 1 && gcd(b1/b0, b1/x) == 1) anstot++;
}
int main () {
    freopen("son.in", "r", stdin);
    freopen("son.out", "w", stdout);
    scanf("%d", &n);
    while (n--) {
        anstot = 0;
        scanf("%d %d %d %d", &a0, &a1, &b0, &b1);
        long long c = sqrt(b1);
        for (int x = 1; x <= c; x++) {
            if (b1 % x != 0) continue;
            check(x);
            if (x * x == b1) continue;
            check(b1 / x);
        }
        printf("%lld\n", anstot);
    }

    return 0;
}