sgu 116. Index of super-prime 背包

最新推荐文章于 2019-06-07 11:08:00 发布

superkunn

最新推荐文章于 2019-06-07 11:08:00 发布

阅读量246

点赞数

CC 4.0 BY-SA版权

分类专栏： sgu !!!!!!!!!! 背包

本文链接：https://blog.youkuaiyun.com/superkunn/article/details/80410162

sgu !!!!!!!!!! 同时被 2 个专栏收录

50 篇文章

订阅专栏

背包

2 篇文章

订阅专栏

本文介绍了一种求解特定正整数作为超级素数之和的最优组合算法。超级素数是指其在素数序列中的位置也是素数的素数。文章详细展示了如何通过动态规划方法找到该组合，并提供了一个C++实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

116. Index of super-prime

time limit per test: 0.25 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

Sample Output

2
3 3

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;


//int v[10005];
int w[201]={3,5,11,17,31,41,59,67,83,109,127,157,179,191,211,241,277,283,331,353,367,401,431,461,509,547,563,587,599,617,709,739,773,797,859,877,919,967,991,1031,1063,1087,1153,1171,1201,1217,1297,1409,1433,1447,1471,1499,1523,1597,1621,1669,1723,1741,1787,1823,1847,1913,2027,2063,2081,2099,2221,2269,2341,2351,2381,2417,2477,2549,2609,2647,2683,2719,2749,2803,2897,2909,3001,3019,3067,3109,3169,3229,3259,3299,3319,3407,3469,3517,3559,3593,3637,3733,3761,3911,3943,4027,4091,4133,4153,4217,4273,4339,4397,4421,4463,4517,4549,4567,4663,4759,4787,4801,4877,4933,4943,5021,5059,5107,5189,5281,5381,5441,5503,5557,5623,5651,5701,5749,5801,5851,5869,6037,6113,6217,6229,6311,6323,6353,6361,6469,6599,6653,6661,6691,6823,6841,6863,6899,7057,7109,7193,7283,7351,7417,7481,7523,7607,7649,7699,7753,7841,7883,8011,8059,8101,8117,8221,8233,8287,8377,8389,8513,8527,8581,8719,8747,8761,8807,8849,8923,8999,9041,9103,9293,9319,9403,9461,9539,9619,9661,9739,9833,9859,9923,9973};
int dp[10005];
int pre[10005];
int ans[10005];
priority_queue<int> pq;
//void SuperPrime(){
//    vector<int> ve;
//    ve.push_back(4);
//    v[0]=1;
//    v[1]=1;
//    for(int i=2;i<=10000;i++){
//        if(v[i]){
//            continue;
//        }
//        ve.push_back(i);
//        for(int j=2*i;j<=10000;j+=i){
//            v[j]=1;
//        }
//    }
//    vector<int>::iterator i;
//    int index=0;
//    for(i=ve.begin();i!=ve.end();i++){
//        if(!v[i-ve.begin()]){
//            cout<<*i<<",";
//            index++;
//        }
//    }
//    cout<<endl;
//    cout<<index<<endl;
//
//}

int main(){
    //SuperPrime();
    for(int i=0;i<201;i++){
        dp[w[i]]=1;
        pre[w[i]]=-1;
        ans[w[i]]=w[i];
    }
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        for(int j=0;j<201;j++){
            if(w[j]>i)break;
            if(dp[i-w[j]]!=0){
                if(dp[i]==0){
                    dp[i]=dp[i-w[j]]+1;
                    pre[i]=i-w[j];
                    ans[i]=w[j];
                }
                else{
                    if(dp[i-w[j]]+1<dp[i]){
                        dp[i]=dp[i-w[j]]+1;
                        pre[i]=i-w[j];
                        ans[i]=w[j];
                    }
                }
            }
        }
    }

    printf("%d\n",dp[n]);
    if(dp[n]==0)return 0;
    for(int i=n;i!=-1;i=pre[i]){
        pq.push(ans[i]);
    }
    while(!pq.empty()){
        printf("%d ",pq.top());
        pq.pop();
    }


    return 0;
}