116 Super-Prime(完全背包+打表预处理)

/*题目大意：

在一个由素数组成的数列 2,3,5,7......

定义超级素数，数列中的第k项为超级素数当且仅当k也为素数

给定n

如果n能被超级素数的和所表示

输出最少要多少个超级素数，以及这些超级素数

否则输出impossible

暴力找出超级素数,跑完全背包即可。*/

 

116. Index of super-prime time limit per test: 0.5 sec.  memory limit per test: 4096 KB Let P1, P2, … ,PN, … be a sequence of prime numbers. Super-prime number is such a prime number that its current number in prime numbers sequence is a prime number too. For example, 3 is a super-prime number, but 7 is not. Index of super-prime for number is 0 iff it is impossible to present it as a sum of few (maybe one) super-prime numbers, and if such presentation exists, index is equal to minimal number of items in such presentation. Your task is to find index of super-prime for given numbers and find optimal presentation as a sum of super-primes. Input There is a positive integer number in input. Number is not more than 10000. Output Write index I for given number as the first number in line. Write I super-primes numbers that are items in optimal presentation for given number. Write these I numbers in order of non-increasing. Sample Input 6 Sample Output 2 3 3

 

/**********************
* author:crazy_石头
* Pro:SGU 116
* algorithm:dp
* Time:15ms
* Judge Status:Accepted
***********************/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>

using namespace std;

#define LL long long
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define INF 1<<29

const int maxn=10000+20;

int prime[maxn],check[maxn],tot=0;
int ans[maxn],cnt=0,dp[maxn],way[maxn];

inline void get_prime()
{
	memset(check,0,sizeof(check));
	for (int i=2;i<maxn;i++)
	{
		if(!check[i])prime[++tot]=i;
		for (int j=i+i;j<maxn;j+=i)
			check[j]=1;
	}
}

inline void init()
{
	get_prime();
	for (int i=2;i<=tot;i++)
		if (check[i]==0)
			ans[++cnt]=prime[i];
	memset(way,0,sizeof(way));
	for (int i=1;i<maxn;i++)
		dp[i]=INF;
	dp[0]=0;
}//预处理；

int main()
{
	init();
	int n;
	scanf("%d",&n);
	//完全背包；
	for(int i =1;i<=cnt;i++)
	{
		for(int j=ans[i];j<=n;j++)
		{
			if (dp[j-ans[i]]+1<dp[j])
			{
				dp[j]=dp[j-ans[i]]+1;
				way[j]=j-ans[i];
			}
		}
	}
	if(dp[n]<INF)
	{
		printf("%d\n",dp[n]);
		int tmp=n;
		for (int i=1;i<=dp[n]-1;i++) {
			printf("%d ",tmp-way[tmp]);
			tmp=way[tmp];
		}
		printf("%d\n",tmp);
	}
	else puts("0");
	return 0;
}