2015 多校赛第三场 1002 (hdu 5317)

最新推荐文章于 2020-07-31 20:08:40 发布

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最新推荐文章于 2020-07-31 20:08:40 发布

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原文链接：http://www.cnblogs.com/names-yc/p/4703085.html

本文介绍了一种关于区间最大GCD查询（RGCDQ）的问题，通过预处理F函数值来优化算法效率。重点在于利用筛法预处理F值，并通过数组记录前缀信息，实现快速查询。

Description

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

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 3

3 5

Sample Output

注意到F值不大。直接用筛法预处理出F值。用数组num[i][j]记录前 i 个数中有多少个 j。

然后暴力。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define maxn 1000005
int t,l,r,F[maxn],num[maxn][10],cnt[10],ans;
void init(){
    for(int i=2;i<=1000000;i++){
        if(F[i]==0){
            for(int j=i<<1;j<=1000000;j+=i)
                F[j]++;
            F[i]=1;
        }
    }
    for(int i=2;i<=1000000;i++){
        for(int j=1;j<=7;j++){
            num[i][j]=num[i-1][j];
            if(F[i]==j) num[i][j]++;
        }
    }
}
int main(){
    init();
    scanf("%d",&t);
    while(t--){
        ans=1;
        scanf("%d%d",&l,&r);
        for(int i=1;i<=7;i++){
            cnt[i]=num[r][i]-num[l-1][i];
            if(cnt[i]>1) ans=i;
        }
        if(cnt[6]&&cnt[3]) ans=max(ans,3);
        if(cnt[6]&&cnt[2]) ans=max(ans,2);
        if(cnt[4]&&cnt[2]) ans=max(ans,2);
        if(cnt[6]&&cnt[4]) ans=max(ans,2);
        printf("%d\n",ans);
    }
    return 0;
}