多校3 1002 RGCDQ

RGCDQ

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 323    Accepted Submission(s): 162

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

2 2 3 3 5

 

 

Sample Output

1 1

 

 

Source

2015 Multi-University Training Contest 3

 

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:   5326  5325  5324  5323  5322 

  1 #include<stdio.h>

  2 #include<string.h>

  3 #include<math.h>

  4 

  5 int a[1000050],b[1000000],k,f[1000050],s[1000050][10];

  6 

  7 int Sieve(int n)

  8 {

  9   a[1]=0;k=0;a[0]=0;

 10            for (int i = 2; i <= n; i++)

 11                 a[i] =1;

 12            for (int i = 2; i <= sqrt(n); i++)

 13            {

 14                if (a[i])

 15                    for (int j = i; j*i <=n; j++)

 16                         a[j * i] = 0;

 17            }

 18            for (int i = 0; i <= n; i++)

 19            {

 20                if (a[i]==1)

 21                {

 22                     k++;

 23                     b[k]=i;

 24                }

 25            }

 26 }

 27 

 28 int gcd(int a,int b) 

 29 { 

 30     if(a<b) 

 31         return gcd(b,a); 

 32     else if(b==0) 

 33         return a; 

 34     else

 35         return gcd(b,a%b); 

 36 } 

 37 

 38 int main()

 39 {

 40     int T;

 41     int i,j,k;

 42     Sieve(1000000);

 43     memset(f,0,sizeof(f));

 44     for(i=2;i<=1000000;i++)

 45     {

 46         int x=i;

 47         k=1;

 48         while(1)

 49         {

 50           if(x==1)

 51           {

 52             break;

 53           }

 54           if(a[x]==1)

 55           {

 56             f[i]++;

 57             break;

 58           }

 59 

 60           if(x%b[k]==0)

 61           {

 62             f[i]++;

 63             while(x%b[k]==0)

 64             {

 65               x=x/b[k];

 66             }

 67           }

 68           k++;

 69         }

 70         //printf("%d ",f[i]);

 71     }

 72     memset(s,0,sizeof(s));

 73     for(i=1;i<=1000000;i++)

 74     {

 75       for(j=1;j<=7;j++)

 76         s[i][j]=s[i-1][j];

 77       s[i][f[i]]++;

 78     }

 79     scanf("%d",&T);

 80     int l,r;

 81     int num[15];

 82     while(T--)

 83     {

 84         memset(num,0,sizeof(num));

 85         scanf("%d %d",&l,&r);

 86         for(i=1;i<=8;i++)

 87           num[i]=s[r][i]-s[l-1][i];

 88         int ma=0;

 89         for(i=10;i>=1;i--)

 90         {

 91           if(num[i]>0)

 92           {

 93             if(num[i]>=2)

 94             {

 95               if(i>ma)

 96                 ma=i;

 97             }

 98             else

 99             {

100               for(j=i-1;j>=1;j--)

101               {

102                 if(num[j]>0)

103                   if(gcd(i,j)>ma)

104                     ma=gcd(i,j);

105               }

106             }

107 

108           }

109         }

110         printf("%d\n",ma);

111     }

112     return 0;

113 }

View Code