hdu 5417 RGCDQ 2015多校联合训练赛-优快云博客

本文链接：https://blog.youkuaiyun.com/firenet1/article/details/47129893

本文详细介绍了RGCDQ算法的定义、原理以及具体实现步骤。通过实例演示了如何解决给定区间内的最大公共约数查找问题，适用于对算法设计和数论问题求解感兴趣的读者。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

RGCDQ

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1283 Accepted Submission(s): 562

Problem 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

maxGCD(F(i),F(j))

(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

Sample Output

1
1

Source

2015 Multi-University Training Contest 3

每个数字最多7个质因数，数量全部预处理。然后看l，r中有几个数字有7,6.。。。1个质因子的，>= 2的就是一个gcd取最大值

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;

int pri[1000001];
int num[1000001];
int sum[1000001][10];
int inf = 1000000;
void init(){
    int ans = 0;
    memset(num,0,sizeof(num));
    for(int i = 2;i <= 1000000; i++){
        if(num[i] == 0){
            for(int j = i;j <= 1000000; j+=i)
                num[j] ++;
        }
    }
    memset(sum,0,sizeof(sum));
    for(int i = 1;i <= inf ;i++){
        for(int j = 1; j <= 7; j++)
            sum[i][j] = sum[i-1][j];
        sum[i][num[i]]++;
    }
}

int main(){
    int t,l,r;
    init();
    int s[10];
    while(scanf("%d",&t)!=EOF){
        while(t--){
            scanf("%d%d",&l,&r);
            for(int i = 1;i <= 7; i++)
                s[i] = sum[r][i] - sum[l-1][i];
            int ans = 1;
            for(int i = 7;i >= 1; i--){
                if(i == 2 && s[2] + s[4] + s[6] > 1)
                    ans = max(ans,2);
                else if(i == 3 && s[3] + s[6] > 1)
                    ans = max(ans,3);
                else if(s[i] > 1)
                    ans = max(ans,i);
            }
            printf("%d\n",ans);
        }
    }
    return 0;

}