· 假设此时已求出标准完全背包，用$f[j]$表示。

· 本题关键在于，由于有个数限制，那么可以强制令当前状态不满足限制，即若最多取$Have[i]$个，强制令其先取$Have[i] + 1$个，那么减去$f[S - (Have[i] + 1)]$即可，当然需用容斥原理来进行加减。

· 代码：

1 #include 
      
        2 #include 
       
         3 #include 
        
          4  5 using namespace std; 6  7 typedef long long LL; 8  9 const int MAXN = 4 + 5;10 const int MAXV = 1e05 + 10;11 12 LL f[MAXV]= {
    0};13 14 const int N = 4;15 16 int W[MAXN];17 int Have[MAXN];18 19 int T;20 21 void Preparation () {22     f[0] = 1;23     for (int i = 1; i <= N; i ++)24         for (int j = W[i]; j <= MAXV - 10; j ++)25             f[j] += f[j - W[i]];26 }27 28 int main () {29     for (int i = 1; i <= N; i ++)30         scanf ("%d", & W[i]);31     scanf ("%d", & T);32 33     Preparation ();34 35     for (int Case = 1; Case <= T; Case ++) {36         int S;37         for (int i = 1; i <= N; i ++)38             scanf ("%d", & Have[i]);39         scanf ("%d", & S);40 41         LL ans = 0;42         int MAX = (1 << N) - 1;43         for (int state = 0; state <= MAX; state ++) {44             int rest = S;45             int cnt = 0;46             for (int k = 1; k <= N; k ++)47                 if (state & (1 << (k - 1)))48                     cnt ++, rest -= (Have[k] + 1) * W[k];49 50             if (rest < 0)51                 continue;52 53             ans -= ((cnt & 1) ? 1 : - 1) * f[rest];54         }55 56         printf ("%lld\n", ans);57     }58 59     return 0; 60 }61 62 /*63 1 2 5 10 264 3 2 3 1 1065 1000 2 2 2 90066 */