题面

Sol

容斥原理+背包

处理出所有金币无限制条件凑成\(j\)元的方案数

考虑计算

\(c\)只有\(4\)种，可以容斥一波

就是无限制的总方案-\(1\)个硬币超出限制的方案+\(2\)个的-\(3\)个的+\(4\)个的

# include 
    
     # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int _(1005);IL int Input(){    RG int x = 0, z = 1; RG char c = getchar();    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}typedef int Arr[_];int c[4], tot, mx, pw[4] = {1, 2, 4, 8}, s[_], d[4][_];ll f[_ * 100] = {1}, ans;int main(RG int argc, RG char *argv[]){    for(RG int i = 0; i < 4; ++i) c[i] = Input();    tot = Input();    for(RG int i = 1; i <= tot; ++i){        for(RG int j = 0; j < 4; ++j) d[j][i] = (Input() + 1) * c[j];        s[i] = Input(), mx = max(mx, s[i]);    }    for(RG int i = 0; i < 4; ++i)        for(RG int j = c[i]; j <= mx; ++j) f[j] += f[j - c[i]];    for(RG int i = 1; i <= tot; ++i){        ans = 0;        for(RG int j = 0; j < 16; ++j){            RG int op = 1, sum = 0;            for(RG int k = 0; k < 4; ++k)                if(j & pw[k]) op = -op, sum += d[k][i];            ans += (sum > s[i]) ? 0 : (1LL * op * f[s[i] - sum]);        }        printf("%lld\n", ans);    }    return 0;}