洛谷1072 hankson的趣味题 数论乱搞 非标准解法

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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：
  
  ∵lcm(x,b0)=b1
  
  
  ∴x∗b0/gcd(x,b0)=b1
  
  
  ∴gcd(x,b0)=x∗b0/b1
  
  
  ∴gcd(b1/b0,b1/x)=1
• 显然x是b1的因数，那么直接从1到(√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;}

好像A了，，，，=。=