多校3- RGCDQ 分类： 比赛 HDU 2015-07-31 10:50 2人阅读 评论(0) 收藏

RGCDQ
Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit

Status

Practice

HDU 5317
Appoint description:
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
2
2 3
3 5
Sample Output
1
1
没有想到所有的数中最大才为7
直接构造二维前缀数组记录在它之前的各个个数的数目；

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <algorithm>
#define RR freopen("output.txt","r",stdout)
#define WW freopen("input.txt","w",stdin)
typedef long long LL;

using namespace std;


const int MAX = 1000010;

int num[MAX];

int dp[MAX][8];

int main()
{
    memset(num,0,sizeof(num));
    memset(dp,0,sizeof(dp));
    for(int i=2; i<MAX; i++)//开始写成（i*i<MAX）了，手残
    {
        if(!num[i])
        {
            num[i]=1;
            for(int j=2; j*i<MAX; j++)
            {
                num[j*i]++;
            }
        }
    }
    dp[2][1]=1;
    for(int i=3; i<MAX; i++)
    {

        for(int j=1; j<=7; j++)
        {
            dp[i][j]=dp[i-1][j];
        }

        dp[i][num[i]]++;
    }

    int L,R;
    int T;
    int ans;
    int s[8];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&L,&R);
        ans=0;
        for(int i=1; i<=7; i++)
        {
            s[i]=dp[R][i]-dp[L-1][i];
        }
        if(s[7]>=2)
        {
            ans=7;
        }
        else if(s[6]>=2)
        {
            ans=6;
        }
        else if(s[5]>=2)
        {
            ans=5;
        }
        else if(s[4]>=2)
        {
            ans=4;
        }
        else if(s[3]>=2)
        {
            ans=3;
        }
        else if(s[6]&&s[3])
        {
            ans=3;
        }
        else if(s[2]>=2)
        {
           ans=2;
        }
        else if(s[6]&&s[2])
        {
          ans=2;
        }
        else if(s[6]&&s[4])
        {
           ans=2;
        }
        else if(s[4]&&s[2])
        {
            ans=2;
        }
        else
        {
           ans=1;
        }
        printf("%d
",ans);
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。