题面如下：

题意简说：

求 ∏ i = l 2 r 2 % ∏ i = l 1 r 1 \prod_{i = l_2}^{r_2} \% \prod_{i = l_1}^{r_1} ∏i=l2r2%∏i=l1r1 是否等于0

思路 or 题解：

r 2 ! / ( l 2 − 1 ) ∣ r 1 ! / ( l 1 − 1 ) ! r_2! / (l_2 - 1)\ |\ r_1!/(l_1-1)! r2!/(l2−1) ∣ r1!/(l1−1)!
等价
r 2 ! ∗ ( l 1 − 1 ) ! = = r 1 ! ∗ ( l 2 − 1 ) ! r_2!* (l_1-1)! == r_1! * (l_2 - 1)! r2!∗(l1−1)!==r1!∗(l2−1)!

判断 a ∣ b a | b a∣b:
只要满足：任意一个素数 p p p， a a a包含 p p p的数量 ≥ \ge ≥ b b b 包含 p p p 的数量
就可以得出 a ∣ b a | b a∣b

官方题解

我们需要用线性筛去优化我们的代码

我们需要用前缀思想和去优化我们的代码

AC代码如下：

const int maxn = 10000009;
int prime[maxn];
bool sf[maxn];
int num = 0;
int a, b, x, y;
void sushu()
{sf[1] = 1;sf[0] = 1;for (int i = 2; i <= maxn; i++){if (!sf[i])prime[++num] = i;for (int j = 1; j <= num; j++){if (i * prime[j] > maxn)break;sf[i * prime[j]] = 1;if (i % prime[j] == 0)break;}}
}
int cal(int n, int p)
{int res = 0;while (n){res += n / p;n /= p;}return res;
}
void solve()
{cin >> a >> b >> x >> y;for (int i = 1; i <= num; i++){if (cal(y, prime[i]) - cal(x - 1, prime[i]) < cal(b, prime[i]) - cal(a - 1, prime[i])){cout << "No\n";return ;}}cout << "Yes\n";
}
int main()
{buff;sushu();int _;cin >> _;while (_--)solve();
}

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