hdu2136 Largest prime factor

Largest prime factor

[b]Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 11238    Accepted Submission(s): 3976
[/b]

[align=left]Problem Description[/align]
Everybody knows any number can be combined by the prime number.

Now, your task is telling me what position of the largest prime factor.

The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc.

Specially, LPF(1) = 0.

 

[align=left]Input[/align]
Each line will contain one integer n(0 < n < 1000000).

 

[align=left]Output[/align]
Output the LPF(n).

 

[align=left]Sample Input[/align]

1
2
3
4
5

 

[align=left]Sample Output[/align]

0
1
2
1
3

题意就是找到这个数n的最大质因子是第几个素数...
OK，题意很简单，范围有100W，怎么搞呢？
基于素筛的思想，先找到100W内每个素数是第几个，前期工作已经搞定，然后我们想，如果该数如果是素数，则直接输出他是第几个素数即可。如果该数是合数，那么我们一定可以把它拆成2个数的乘积，OK，这里需要一点dp的思路，我们首先假设n=a*b这里，假设我们已经知道了a和b的最大质因子是第几个素数了。下面即可得出状态转移方程
num
=max(num
,max(num[a],num[b]))；n=a*b，即b=n/a；
这样，我们即可在nlogn的范围内找到该问题的解。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<stack>
#include<queue>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#define eps 1e-8
#define zero(x) (((x>0?(x):-(x))-eps)
#define mem(a,b) memset(a,b,sizeof(a))
#define memmax(a) memset(a,0x3f,sizeof(a))
#define pfn printf("\n")
#define ll __int64
#define ull unsigned long long
#define sf(a) scanf("%d",&a)
#define sf64(a) scanf("%I64d",&a)
#define sf264(a,b) scanf("%I64d%I64d",&a,&b)
#define sf364(a,b,c) scanf("%I64d%I64d%I64d",&a,&b,&c)
#define sf2(a,b) scanf("%d%d",&a,&b)
#define sf3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define sf4(a,b,c,d) scanf("%d%d%d%d",&a,&b,&c,&d)
#define sff(a) scanf("%f",&a)
#define sfs(a) scanf("%s",a)
#define sfs2(a,b) scanf("%s%s",a,b)
#define sfs3(a,b,c) scanf("%s%s%s",a,b,c)
#define sfd(a) scanf("%lf",&a)
#define sfd2(a,b) scanf("%lf%lf",&a,&b)
#define sfd3(a,b,c) scanf("%lf%lf%lf",&a,&b,&c)
#define sfd4(a,b,c,d) scanf("%lf%lf%lf%lf",&a,&b,&c,&d)
#define sfc(a) scanf("%c",&a)
#define ull unsigned long long
#define pp pair<int,int>
#define debug printf("***\n")
const double PI = acos(-1.0);
const double e = exp(1.0);
const int INF = 0x7fffffff;;
template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<class T> inline T Min(T a, T b) { return a < b ? a : b; }
template<class T> inline T Max(T a, T b) { return a > b ? a : b; }
bool cmpbig(int a, int b){ return a>b; }
bool cmpsmall(int a, int b){ return a<b; }
using namespace std;
int heshu[1000010];
int num[1000010];
int sushu[1000010];
int main()
{
//freopen("data.in","r",stdin);
mem(heshu,0);
mem(sushu,0);
mem(num,0);
int i,j;
heshu[0]=heshu[1]=1;
for(i=2;i*i<=1000000;i++)
for(j=i;j*i<=1000000;j++)
heshu[i*j]=1;
int flag=0;
for(i=1;i<=1000000;i++)
{
if(heshu[i]==0)
{
sushu[i]=++flag;
num[i]=flag;
}
}
for(i=2;i*i<=1000000;i++)
for(j=i;j*i<=1000000;j++)
num[i*j]=max(num[i*j],max(num[i],num[j]));
int n;
while(~sf(n))
{
printf("%d\n",num
);
}
return 0;
}