看成01背包即可

数量是多少就存储多少次！

写法一

#include <iostream>
using namespace std;
const int N = 10010;
int weight[N], value[N], f[N];
int main(){
    int n, bag;
    cin >> n >> bag;
    int k = 1;
    int w, v, num;
    for(int i = 1; i <= n; i ++ ){
        cin >> w >> v >> num;
        for(int j = 1; j <= num; j ++ ){
            weight[k] = w, value[k ++ ] = v;
        }
    }
    for(int i = 1; i < k; i ++ ){
        for(int j = bag; j >= weight[i]; j -- ){
            f[j] = max(f[j], f[j - weight[i]] + value[i]);
        }
    }
    cout << f[bag] << endl;
    return 0;
}

写法二

#include <iostream>
using namespace std;
const int N = 110;
int weight[N], value[N], amount[N]; 
int f[N];
int main(){
    int n, bag;
    cin >> n >> bag;
    for(int i = 1; i <= n; i ++ )   cin >> weight[i] >> value[i] >> amount[i];
    for(int i = 1; i <= n; i ++ ){
        for(int j = bag; j >= weight[i]; j -- ){
            for(int k = 0; k * weight[i] <= j && k <= amount[i]; k ++ ){
                f[j] = max(f[j], f[j - k * weight[i]] + k * value[i]);
            }
        }
    }
    cout << f[bag] << endl;
    return 0;
}