hdu 5317 RGCDQ (소수 체 + 전달 방정식)
문제 면 설명:
RGCDQ
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2825 Accepted Submission(s): 1111
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
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
1 1
题目大意:
定义:F(x)表示x可以被素因子分解得到的素因子的种类数,给定区间[L,R],问区间内任意两个F(i)和F(j)之间的最大公约数的最大值为多少。
题目分析:
第一步肯定要先处理出来F(x)函数(肯定是通过素数筛,素因子分解的方法),然后再思考如果每次都在给定区间内暴力寻找,这样的方法肯定会超时。
通过大表和计算发现,由于2*3*5*7*11*13*17*19=9699690>1000000,所以,1000000以内的所有的数的素因子的种类数都小于等于7,所以,我们借助于地推的思想,记sum[i][j] 为从2到i区间内含有j种素因子数的个数,则递推关系为:sum[i][j]=sum[i-1][j]+1;(如果F(j)==j); 或sum[i][j]=sum[i][j-1];(F[j]!=j);
那么区间L到R的含有i个素因子的个数就等于sum[R][i]-sum[L-1][i]个,这样如果存在大于等于两个素因子种类数的个数为一样的值,那么最大的最大公约数就是该值,但是也要考虑一下特殊情况,比如同时有2,4,6(4,6)或者3,6的时候,取一下最大值就行了。
代码实现:
#include
#include
#include
using namespace std;
const int MAXN=1000100;///MAXN
int prime[MAXN+1];
void getPrime()
{
memset(prime,0,sizeof(prime));
for(int i=2;i<=MAXN;i++)
{
if(!prime[i])
prime[++prime[0]]=i;
for(int j=1;j<=prime[0]&&prime[j]<=MAXN/i;j++)
{
prime[prime[j]*i]=1;
if(i%prime[j]==0)
break;
}
}
}
long long factor[1000100];
int fatCnt;
int getFactors(long long x)
{
fatCnt=0;
long long tmp=x;
for(int i=1;prime[i]<=tmp/prime[i];i++)
{
if(tmp%prime[i]==0)
{
factor[fatCnt]=prime[i];
while(tmp%prime[i]==0)
{
tmp/=prime[i];
}
fatCnt++;
}
}
if(tmp!=1)
{
factor[fatCnt++]=tmp;
}
return fatCnt;
}
int F[1000010];
int sum[1000010][10];
int main()
{
int T,L,R;
getPrime();
memset(F,0,sizeof(F));
memset(sum,0,sizeof(sum));
for(long long i=2;i<=1000000;i++)
{
int flag=getFactors(i);
F[i]=flag;
}
for(int i=2;i<=1000000;i++)/// 2-i(2<=i<=1000000) j
{
for(int j=1;j<=7;j++)
{
if(F[i]==j)
{
sum[i][j]=sum[i-1][j]+1;
}
else
{
sum[i][j]=sum[i-1][j];
}
}
}
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&L,&R);
int dis[10];
memset(dis,0,sizeof(dis));
for(int i=1;i<=7;i++)
{
dis[i]=sum[R][i]-sum[L-1][i];
}
int ans=1;
for(int i=7;i>=1;i--)
{
if(dis[i]>=2)
{
ans=i;
break;
}
}
if((dis[2]>0) && (dis[4]>0||dis[6]>0))
ans=max(ans,2);
if(dis[3]>0 && dis[6]>0)
ans=max(ans,3);
if(dis[4]>0 && dis[6]>0)
ans=max(ans,2);
printf("%d
",ans);
}
return 0;
}
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