SGU116 动态规划 DP

问题:超级质数是在质数序列中下标也是质数的质数,例如3,5,11……现在给你一个数,找一个由超级质数构成的最小表示使得这些数的和等于这个数。

解法:先求出这些超级质数,然后就是个完全背包问题。具体可见我动态规划关于背包的文章。

Problem: Super-prime number is such a prime number that its current number in prime numbers sequence is a prime number too. For example, 3,5,11...Your task is to find index of super-prime for given numbers and find optimal presentation as a sum of super-primes.

Solution: At first, calculate all the super-prime you need, then it comes into a complete knapsack problem. You can find more about the topic in my articles.

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define inf 1010001000
int cnt,pnum,p[10000],q[10000],n,dp[10100],his[10100],k,ans[10100];
bool f[10010];
void output(int x) {
     if (x<=0) return;
     ans[++k] = x-his[x];
     output(his[x]);
}
int main() {
    pnum = 0;
    memset(f,0,sizeof(f));
    for (int i=2;i<=10000;i++)
        if (!f[i]) {
           p[++pnum] = i;
           for (int j=i;j<=10000;j+=i) f[j] = true;
        }
    cnt = 0;
    for (int i=1;i<=pnum;i++)
        if (p[i]<=pnum) q[++cnt] = p[p[i]];
    
    while (~scanf("%d",&n)) {
          for (int i=1;i<=10010;i++) dp[i] = -inf;
          dp[0] = 0;
          for (int i=1;i<=cnt;i++)
              for (int j=q[i];j<=n;j++)
                  if (dp[j]<dp[j-q[i]]-1) {
                     dp[j] = dp[j-q[i]]-1;
                     his[j] = j-q[i];
                  }
          if (dp[n]==-inf) printf("0\n");
          else {
               printf("%d\n",-dp[n]);
               k = 0;
               output(n);
               sort(ans+1,ans+k+1);
               for (int i=k;i>1;i--) printf("%d ",ans[i]);
               printf("%d\n",ans[1]);
          }
    }
    return 0;
}


SGU116 动态规划 DP

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