hdu 5317 RGCDQ（数论素筛）

RGCDQ

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

 

Author

ZSTU

 

Source

2015 Multi-University Training Contest 3

参考博客：http://www.cnblogs.com/Ritchie/p/5312564.html

ps：
一看题意如果按照常规方法写的话，就肯定会超时的，所以我们可以先对f[i]打个表，观察一下它的规律

我们可以发现2*3*5*7*11*13*17=510510，2*3*5*7*11*13*17*19=9699690，也就是说100w以内的数的素因数种类最多为7

因此，我们可以先预处理打表，在素数筛法的过程中，让a[i]表示i的素因子种类数，让b[i][j]表示2到i中素因子种类数为j的有多少个

打完表后就可以让i从7到1开始遍历，只要有b[R][i]-b[L-1][i]>=2（就说明在L~R区间内素因子种类数为i的数至少有两个），就令ans=i，并且break输出即可（因为我们要的是最大值）

代码：

#include<stdio.h>
#include<string.h>

#define maxn 1000000+10
int a[maxn],b[maxn][8];

void pre_table()
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(int i=2; i<maxn; i++)
{
if(!a[i])
for(int j=i+i; j<maxn; j+=i)
a[j]++;
}
for(int i=2;i<maxn;i++)
for(int j=1;j<=7;j++)
{
b[i][j]=b[i-1][j];
if(a[i]==j)
b[i][j]++;
}
}

int main()
{
pre_table();
int t;
scanf("%d",&t);
while(t--)
{
int L,R;
scanf("%d%d",&L,&R);
int ans=1;
for(int i=7;i>=1;i--)
{
if(b[R][i]-b[L-1][i]>=2)
{
ans=i;
break;
}
}
printf("%d\n",ans);
}
return 0;
}

总结：
1.学到了在素数筛法中记录合数素因子种类数的方法
2.一开始我也打表看了看，但是没有想到用一个二维数组记录区间内的素因子种类数，所以这一点的思维性还是欠缺的0.0