HDU5317：RGCDQ （数学 & 二分）

最新推荐文章于 2017-11-10 08:54:44 发布

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本文介绍了一种算法挑战，即区间质因子GCD查询问题。任务是在给定的区间内找出两个数的最大质因子数量的GCD。通过预处理所有数的质因子数量，并使用二分查找来高效解答多个询问。

Mr. Hdu is interested in Greatest Common Divisor (GCD). He wants to find more and more interesting things about GCD. Today He comes up with Range Greatest Common Divisor Query (RGCDQ). What’s RGCDQ? Please let me explain it to you gradually. For a positive integer x, F(x) indicates the number of kind of prime factor of x. For example F(2)=1. F(10)=2, because 10=2*5. F(12)=2, because 12=2*2*3, there are two kinds of prime factor. For each query, we will get an interval [L, R], Hdu wants to know maxGCD(F(i),F(j))maxGCD(F(i),F(j)) (L≤i<j≤R)(L≤i<j≤R)

Input

There are multiple queries. In the first line of the input file there is an integer T indicates the number of queries.
In the next T lines, each line contains L, R which is mentioned above.

All input items are integers.
1<= T <= 1000000
2<=L < R<=1000000

Output

For each query，output the answer in a single line.
See the sample for more details.

Sample Input

2
2 3
3 5

Sample Output

1
1

题意：f[i]表示i的不同质因子数。T个询问，输出区间内gcd(f[i],f[j])最大的值。

思路：1e6范围内质因子至多7个不同，打个表对每个数的质因子数存下，二分查找即可。或者用前缀和也可以。

# include <iostream>
# include <cstdio>
# include <set>
# include <algorithm>
using namespace std;
const int N = 1e6+30;
int isprime[N+3], pcnt, cnt[N+3];
vector<int>v[10];
int main()
{
    int t, l, r;
    for(int i=2; i<N; ++i)
    {
        if(!isprime[i])
        {
            for(int j=i; j<N; j+=i)
            {
                isprime[j] = 1;
                ++cnt[j];
            }
        }
    }
    for(int i=2; i<=N-30; ++i)
        v[cnt[i]].push_back(i);
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            scanf("%d%d",&l,&r);
            int i;
            for(i=8; i>=1; --i)
            {
                int p1 = upper_bound(v[i].begin(), v[i].end(), r)-v[i].begin()-1;
                int p2 = lower_bound(v[i].begin(), v[i].end(), l)-v[i].begin();
                if(p1-p2+1 >= 2)
                {
                    printf("%d\n",i);
                    break;
                }
            }
            if(i==0) puts("1");
        }
    }
    return 0;
}