[HAOI2008]硬币购物

· 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：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

typedef long long LL;

const int MAXN = 4 + 5;
const int MAXV = 1e05 + 10;

LL f[MAXV]= {0};

const int N = 4;

int W[MAXN];
int Have[MAXN];

int T;

void Preparation () {
    f[0] = 1;
    for (int i = 1; i <= N; i ++)
        for (int j = W[i]; j <= MAXV - 10; j ++)
            f[j] += f[j - W[i]];
}

int main () {
    for (int i = 1; i <= N; i ++)
        scanf ("%d", & W[i]);
    scanf ("%d", & T);

    Preparation ();

    for (int Case = 1; Case <= T; Case ++) {
        int S;
        for (int i = 1; i <= N; i ++)
            scanf ("%d", & Have[i]);
        scanf ("%d", & S);

        LL ans = 0;
        int MAX = (1 << N) - 1;
        for (int state = 0; state <= MAX; state ++) {
            int rest = S;
            int cnt = 0;
            for (int k = 1; k <= N; k ++)
                if (state & (1 << (k - 1)))
                    cnt ++, rest -= (Have[k] + 1) * W[k];

            if (rest < 0)
                continue;

            ans -= ((cnt & 1) ? 1 : - 1) * f[rest];
        }

        printf ("%lld
", ans);
    }

    return 0; 
}

/*
1 2 5 10 2
3 2 3 1 10
1000 2 2 2 900
*/