如何转换成背包问题呢，我们可以把每个质数当成一个重量

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;#define int long long
int record[1005];
void fun() {//record[2] = 1;for (int i = 2; i <= 1000; i++) {if (!record[i]) {for (int j = i * 2; j <= 1000; j += i) record[j] = 1;}}
}int n;
int dp[1005];signed main() {fun();cin >> n;dp[0] = 1;for (int i = 2; i <= n; i++) {  // 物品的重量if (record[i])  continue;for (int j = i; j <= n; j++) {dp[j] += dp[j - i];}}cout << dp[n];
}

再看一题，这种有容量限制的都可以转换为背包问题

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;const int N = 1005;
int va[N];
int is[N];
int n;
void fun() {va[2] = 1;for (int i = 3; i <= n; i++) {va[i] += 1;for (int j = 2; j < i; j++) {if (i % j == 0) {va[i] += j;//if (i / j != j) {//	va[i] += i / j;//}}}}//for (int i = 2; i <= 10; i++) {//	cout << i << " " << va[i] << endl;//}
}
int dp[N];int main() {cin >> n;fun();for (int i = 1; i <= n; i++) {for (int j = i; j <= n; j++) {dp[j] = max(dp[j], dp[j - i] + va[i]);}}cout << dp[n];
}