EOJ月赛-2021.4 --D. Divide(枚举质因子)

题目链接

题目思路：

预处理1-1e7的所有质因子

然后对区间L,R

我们算出每个质因子的个数

上述操作重复两边（两个区间）

然后check的时候对于所有的质因子是不是都满足num1[i]<=num2[i]就可以了

ll vis[10000002], p[10000002], x;
void oula()
{
    for(int i = 2 ; i <= 10000000 ; i++)
    {
        if(vis[i] == 0) p[++x] = i;
        for(int j = 1 ; j <= x && p[j]*i <= 10000000; j++)
        {
            vis[i * p[j]] = 1;
            if(i % p[j] == 0) break;
        }
    }
}
ll l1, r1, l2, r2, num1[10000002], num2[10000002];
void get1()
{

    for(int i = 1 ; i <= x ; i++)
    {
        ll t = p[i];
        if(t > r1) break;
        ll sum = 0 ;
        while(t <= r1)
        {
            int s1 = r1 / t;
            int s2 = (l1 - 1) / t;
            sum += s1 - s2;
            t = t * p[i];
        }
        num1[i] = sum;
    }
}
void get2()
{
    for(int i = 1 ; i <= x ; i++)
    {
        ll t = p[i];
        if(t > r1) break;
        ll sum = 0 ;
        while(t <= r2)
        {
            ll s1 = r2 / t;
            ll s2 = (l2 - 1) / t;
            sum += s1 - s2;
            t = t * p[i];
        }
        num2[i] = sum;
    }
}
int ok()
{
    for(int i = 1 ; i <= x ; i++) if(num1[i] > num2[i] ) return 0;
    return 1;
}
int main()
{
    oula();
    int _ = read();
    while(_--)
    {
        mst(num1, 0), mst(num2, 0);
        l1 = read(), r1 = read(), l2 = read(), r2 = read();
        get1(), get2();
        if(ok()) puts("Yes");
        else puts("No");
    }
    return  0 ;
}

标签：月赛,return,2021.4,--,ll,int,while,read,r1
来源： https://www.cnblogs.com/UpMing/p/14642327.html