hdoj5317RGCDQ

RGCDQ

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2711    Accepted Submission(s): 1070

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

22 33 5

 

Sample Output

11

 

Author

ZSTU

 

Source

2015 Multi-University Training Contest 3

 

解题思路：打出质因子表记录前缀和因为所给范围内的数的不同的质因子的个数最大为7

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<map>#include<stack>using namespace std;const int maxn=1000010;int f[maxn];int pre[maxn][10];bool vis[maxn],used[10];int gcd(int a,int b){    return b==0?a:gcd(b,a%b);}void dabiao(){    for(int i=2;i<maxn;++i){        if(vis[i])continue;vis[i]=true;        for(int j=i;j<maxn;j+=i){            f[j]++;vis[j]=true;        }    }    for(int i=2;i<maxn;++i){        for(int j=1;j<=7;++j){            if(f[i]==j){                pre[i][j]=pre[i-1][j]+1;            }            else {                pre[i][j]=pre[i-1][j];            }        }    }}int main(){    dabiao();    int t;cin>>t;    while(t--){        int l,r;        scanf("%d%d",&l,&r);        int ans=0;        memset(used,false,sizeof(used));        for(int i=1;i<=7;++i){            int cnt=pre[r][i]-pre[l-1][i];            if(cnt>1){                used[i]=true;                ans=max(ans,i);            }            else if(cnt==1){                used[i]=true;            }        }        for(int i=1;i<=7;++i){            for(int j=i+1;j<=7;++j){                if(used[i]&&used[j]){                    ans=max(ans,gcd(i,j));                }            }        }        printf("%d\n",ans);    }    return 0;}