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
==-inf) printf("0\n");
else {
printf("%d\n",-dp
);
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;
}