#include [HTML_REMOVED]
#include [HTML_REMOVED]
#include [HTML_REMOVED]

using namespace std;

typedef long long i64;

const int N = 245000;

int n, p;
i64 m;
int s[N], c[N];
i64 f[N];

int main() {
cin >> n >> m;
for (int i = 1; i <= n; i ) cin >> s[i];
for (int i = 1; i <= n; i ) cin >> c[i], p += s[i] * c[i];

f[0] = 1;
for (int i = 1; i <= n; i ++ ) 
    for (int j = p; j >= c[i]; j -- )
        for (int k = 1; k * c[i] <= j && k <= s[i]; k ++ )
            f[j] = max(f[j], f[j - k * c[i]] * k);

for (int i = 0; i <= p; i ++ )
    if (f[i] >= m) {
        cout << i;
        break;
    }

return 0;

佬，为什么要先枚举体积再枚举个数呢

为什么先枚举个数会出错呢

要先装得下，才会考虑装多少个

对啊，而且这个题当成多重背包做，就是这个代码我不知道为什么会有问题

/*

由于花费较少，所以用 f[i][j] 代表前 i 件物品，不超过 j 所能达到的最大策略数

f[i][j] = max(f[i - 1][j], f[i - 1][j - k * c[i]] * k;

*/

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 4e5 + 10, M = 510;

int s[M], c[M];

long long n, m;

long long f[N];

int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i ++) cin >> s[i];
    for(int i = 1; i <= n; i ++) cin >> c[i];

    f[0] = 1;
    for(int i = 1; i <= n; i ++)
    {
        for(int k = 1; k <= s[i]; k ++)
        {
            for(int j = 2e5 + 10; j >= k * c[i]; j --)
            {
                f[j] = max(f[j], f[j - k * c[i]] * k);
            }
        }
    }
    long long ans = 1e18;
    for(int i = 1; i <= N; i ++)
    {
        if(f[i] >= m) 
        {
            ans = min(ans, (long long)i);  
            //cout << f[i] << endl;

        }//cout << f[n & 1][i] << endl;
    }
    cout << ans << endl;
    return 0;
}

错了，不是这样的